Explanation

Let electric filed be E
Water drop is subjected to upward electric force FE=eE
Down word gravitational force FG=mg
To keep water drop suspended resultant force must be zero
FE=FG
On substituting values of FE and FG
eE=mg
E=mg/e

Explanation

Both the capacitors are in series and connected across the source of 500V. Hence charge on the each capacitor will be same.Effective capacitance

On substituting the values of both the capacitors , after solving for C, we get c=2.4μF
Now Q=CV
Q=2.4μF ×500V
Q=1200μC

Explanation

Function of capacitor is to store energy in the form of electric field and charge in accumulated in it when connected to high voltage source.
When such a charged capacitor is conned to resistors or any conductor charge flows through it.
Our body is a conductor of electric charge. When we touch capacitor, it tries to send out charge through our body to ground. Which can give fatal shock to the body if discharge current is high.

Explanation:

Electrostatics is the branch of physics that deals with electric charges at rest and the electric field produced by stationary charges.

If charges are not moving:
Current I = 0
Magnetic effects due to current do not appear.

Electromagnetism is broader and includes both electric and magnetic fields, especially when charges are moving or fields are changing with time.

Magnetostatics specifically deals with magnetic fields produced by steady (constant) currents, not electric charges at rest.

Therefore, the study of the effect associated with an electric field at rest is called electrostatics.

Explanation

All the capacitors are connected in series hence charge on each capacitor is same

Equivalent capacitor On substituting the value and taking reciprocal
we get C=1 μF
Now Q=CV
Therefore Q=1 μF × 10V
Q=10 μ C

Explanation

Given Coulomb's Law electrostatic forces: F=kQq / d²

We can see that distance and force are inversly related. Also distance is squared, so if we increase the distance by 2, the force between the two charges will be reduced by a factor of four.

Explanation:

A positive test charge will naturally move from high potential to low potential. If it is moved in the opposite direction, then the electric field will do work against its motion (negative work). This be seen from the equation for electric field work:

W=−qΔV

W is the work done by the electric field, q is the charge, and ΔV is the potential difference. If ΔV is positive (the final potential is higher than the initial potential) and q is also positive, then work done by the field is negative.

Explanation:

Electrical potential energy is given by the equation

Electrical potential energy is inversely proportional to the distance between the two charges. If the energy is quadrupled, then r (the distance between the two equal charges) must have decreased proportionally.

Explanation:

The resistor will absorb power and dissipate it in the form of heat energy. The potential between two points across a resistor will be negative.

Current flows from higher potential to lower potential through a resistor.
So in the direction of current, potential continuously drops.
Therefore (Vfinal − Vinitial) along current direction is negative.

Explanation:

A capacitor is made of two conducting plates separated by an insulating material (dielectric).

So it consists of two conductors.

Explanation:

Capacitive reactance:
Xc = 1 / (2π f C)

When frequency f becomes very small (f → 0),
Xc becomes very large (→ ∞).

So at very low frequencies (and at DC), a capacitor blocks current and behaves like an open circuit.

Explanation:

Capacitive impedance is inversely proportional to frequency. Hence at very high frequencies, the impedance is almost equal to zero, hence it acts as a short circuit and there is no voltage across it.

Capacitive reactance:
Xc = 1 / (2π f C)

When frequency f increases, Xc decreases and approaches zero.

So at high frequencies, a capacitor offers very small opposition and behaves like a short circuit.

Explanation:

Charge stored in a capacitor:
Q = C V

Given
C = 3 F
V = 2 V

Q = 3 × 2
Q = 6 C

Explanation:

The current through a capacitor depends on how fast the voltage across it is changing.
If voltage changes quickly, more current flows; if voltage is constant, current becomes zero.
So the correct relation is I = C (dV/dt).

Explanation:

In a DC charging circuit, the current gradually decreases as the capacitor voltage increases.
At the final stage, capacitor voltage becomes equal to the source voltage, so potential difference across resistor becomes zero.
Therefore, final charging current becomes 0 A.

Explanation:

When capacitor voltage becomes equal to source voltage, the potential difference across the resistor becomes zero.
So current becomes zero, meaning the capacitor stops taking more charge.
Now the capacitor is fully charged and behaves like an open circuit for DC.

Explanation:

In an RC charging circuit, the capacitor voltage increases with time and approaches the supply voltage.

A capacitor is considered fully charged (steady state) when:
Vc = Vs

At that moment, there is no potential difference across the resistor, so charging current becomes zero.

Explanation:

I(t) = (V₀/R) e^−t/RC

As t → ∞,
e^−t/RC → 0

So,
I_final = 0 A

Explanation:

For a capacitor discharging through a resistor R, the initial discharge current (at t = 0) is

I₀ = V₀ / R

Given
V₀ = 400 V
R = 20 Ω

I₀ = 400 / 20
I₀ = 20 A

Explanation:

Force between two point charges has the same magnitude on each charge (Newton’s third law).

Using Coulomb’s law (magnitude):
F = k |q1 q2| / d²

This F is the magnitude of force on +4 mC due to −1 mC.
The force on −1 mC due to +4 mC has the same magnitude, opposite direction.

So the ratio of forces acting on them is:
F on (+4 mC) : F on (−1 mC) = 1 : 1

Explanation:

In an RC charging circuit, at the start (t = 0), the capacitor is uncharged, so it behaves like a short circuit.

Initial charging current:
I₀ = V / R

Given
V = 400 V
R = 20 Ω

I₀ = 400 / 20
I₀ = 20 A

Explanation:

In an RC circuit, as time becomes very large (t → ∞), the capacitor becomes fully charged.

Then capacitor voltage becomes equal to battery voltage, so no potential difference remains across the resistor.

So current:
I = 0

Explanation:

In an RC charging circuit, at the instant charging starts (t = 0), the capacitor is uncharged.

So capacitor voltage:
Vc(0) = 0

That means the whole battery voltage appears across the resistor, so current is maximum:

I(0) = V / R

After that, current decreases as the capacitor charges.

So the initial current is high (maximum).

Explanation:

In an RC circuit, the charging and discharging rate is governed by the time constant.

Time constant:
τ = R C

Larger τ means slower charging/discharging.
Smaller τ means faster charging/discharging.

So the quantity that determines (depends on) the charging and discharging rate is the time constant.

Explanation

Explanation:

Each capacitor has capacitance 2 mF.

Check each arrangement.

Option A: all three in series
1/Ceq = 1/2 + 1/2 + 1/2 (mF)⁻¹
1/Ceq = 3/2
Ceq = 2/3 mF
Not 3 mF.

Option B: all three in parallel
Ceq = 2 + 2 + 2
Ceq = 6 mF
Not 3 mF.

Option C: two in series, then third in parallel with that combination
First two in series:
Cseries = (2 × 2) / (2 + 2)
Cseries = 4/4
Cseries = 1 mF

Now in parallel with third (2 mF):
Ceq = 1 + 2
Ceq = 3 mF
Matches required.

Option D: two in parallel, then third in series with that combination
First two in parallel:
Cparallel = 2 + 2
Cparallel = 4 mF

Now series with third (2 mF):
Ceq = (4 × 2) / (4 + 2)
Ceq = 8/6
Ceq = 4/3 mF
Not 3 mF.

Explanation:

For capacitors in series, the same charge Q appears on each capacitor and the total potential is the sum:

V = V1 + V2

Using V = Q/C:
Q/Ceq = Q/C1 + Q/C2

Divide by Q:
1/Ceq = 1/C1 + 1/C2

So
Ceq = C1C2 / (C1 + C2)

Explanation:

A charged capacitor stores energy in its electric field.

Energy density of electric field:
u = 1/2 ε E²

This energy is distributed throughout the region where the electric field exists, mainly between the plates (fringing at edges is small compared to the main region).

Therefore, energy resides in the field between the plates.

Explanation:

When the electric field (potential gradient) in a dielectric becomes so large that the dielectric breaks down (gets punctured), conduction starts through it.
The maximum electric field a dielectric can withstand without breakdown is called dielectric strength.

Explanation:

A metal foil is a conductor, so the electric field inside the foil is zero and the foil becomes an equipotential surface.

If the foil is inserted between the plates without touching them, the capacitor becomes equivalent to two capacitors in series:
One between left plate and foil (separation d₁)
One between foil and right plate (separation d₂)

Total separation is still
d₁ + d₂ = d

Equivalent capacitance:
1/Ceq = 1/C1 + 1/C2
C1 = εA/d₁
C2 = εA/d₂

So
1/Ceq = d₁/(εA) + d₂/(εA)
1/Ceq = (d₁ + d₂)/(εA)
= d/(εA)

Therefore
Ceq = εA/d

This is the same as the original capacitance.

Explanation:

W = U = Q² / (2C)

Q = 8 × 10⁻¹⁸ C
C = 100 μF = 100 × 10⁻⁶ F = 1 × 10⁻⁴ F

Q² = (8 × 10⁻¹⁸)²
Q² = 64 × 10⁻³⁶

W = (64 × 10⁻³⁶) / (2 × 10⁻⁴)
W = (64/2) × 10^{−36 − (−4)}
W = 32 × 10⁻³² J

Explanation:

A capacitor stores charge and energy.

It can deliver a large amount of charge in a very short time, so it is used where a device needs a sudden extra surge of current (for example at starting).

So the correct choice is current.

Explanation:

Energy stored in a capacitor is
U = 1/2 C V²

So it depends on the square of the potential.

Explanation:

Let the first conductor have capacitance C1 and charge Q.

Initial energy:
U1 = Q² / (2C1)

Connect it to an uncharged conductor of capacitance C2.
After connection, charge redistributes but total charge remains Q.

The two conductors effectively form a system with larger total capacitance:
Ctotal = C1 + C2

Final energy:
U2 = Q² / (2Ctotal)
U2 = Q² / (2(C1 + C2))

Since (C1 + C2) > C1, the denominator is larger, so
U2 < U1

The decrease in energy appears as heat (and a little radiation) during charge redistribution.

Explanation:

For a capacitor fully filled with a dielectric:

C₀ (air/vacuum) = ε₀ A / d
With dielectric, ε = K ε₀

New capacitance:
C = ε A / d
C = (K ε₀) A / d
C = K (ε₀ A / d)
C = K C₀

So capacitance increases K times.

Explanation:

For a parallel plate capacitor connected to a battery, the battery keeps the potential difference constant.

So
V = constant

When a dielectric of constant K completely fills the space:
C₀ = ε₀ A / d
C = K ε₀ A / d
C = K C₀

So capacitance increases.

Charge on capacitor:
Q = C V

Since V is constant and C increases, therefore
Q increases.

Electric field between plates:
E = V / d

Since V is constant and d is constant, therefore
E remains constant (does not increase).

So the quantities that increase are Q and C.

Explanation:

Explanation:

For a parallel plate capacitor:

C = ε A / d

Here A is plate area, ε is permittivity, and d is separation.

If d decreases, the denominator becomes smaller, so C increases.

Explanation:

On an equipotential surface, potential is the same at every point.

So potential difference:
ΔV = 0

Work done in moving a charge q is:
W = q ΔV

For a unit positive charge, q = 1 C:
W = 1 × 0
W = 0

So no work is done

Explanation:

Statement I
Potential difference between two points in an electrostatic field is path independent.
So it depends only on initial and final points.
This is true.

Statement II
Definition of potential difference:
ΔV = W/q
So it is the work done per unit positive test charge in moving from one point to another.
This is true.

Statement III
Potential difference is W/q.
If charge is doubled, work done also doubles in the same field, so W/q remains the same.
So potential difference is not more for a 2-unit charge than for a 1-unit charge.
This is false.

Therefore correct statements are I and II only.