Electrostatic Potential and Capacitance - Live Session - NEET 2020Contact Number: 9667591930 / 8527521718

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The potential of the electric field produced by a point charge at any point (x, y, z) is given by $V=3{x}^{2}+5$, where x, y, z are in metres and V is in volts. The intensity of the electric field at (–2, 1, 0) is

1. +17 Vm^{–1}

2. –17 Vm^{–1}

3. +12 Vm^{–1 }

4. –12 Vm^{–1}

Eight charged water drops each with a radius of 1mm and a charge of 10^{–10} C merge into single drop. Potential of the big drop will be

1. 3600 V

2. 900 V

3. 300 V

4. 150 V

Three concentric spherical shells have radii a, b and

c (a < b < c) and have surface charge densities $\mathrm{\sigma}$, –$\mathrm{\sigma}$

and $\mathrm{\sigma}$ respectively. If V_{A}, V_{B} and V_{C} denote the

potentials of the three shells, then for c = a + b, we

have

1. VC = VB $\ne $ VA

2. VC $\ne $ VB $\ne $ VA

3. VC = VB = VA

4. VC = VA $\ne $ VB

Charges +q and –q are placed at points A and B respectively which are at a distance 2L apart, C is the mid-point between A and B. The work done in moving a charge +Q along the semicircle CRD is

1. $\frac{\mathrm{qQ}}{2{\mathrm{\pi \epsilon}}_{0}\mathrm{L}}$

2. $\frac{\mathrm{qQ}}{6{\mathrm{\pi \epsilon}}_{0}\mathrm{L}}$

3. $\frac{-\mathrm{qQ}}{6{\mathrm{\pi \epsilon}}_{0}\mathrm{L}}$

4. $\frac{\mathrm{qQ}}{4{\mathrm{\pi \epsilon}}_{0}\mathrm{L}}$

An air capacitor C, is connected to a battery of emf V. It acquires a charge Q and energy E. The capacitor is then disconnected from the battery and a dielectric slab is introduced between the plates. Which of the following is true?

1. V & Q decrease but E & C increase

2. V remains unchanged but Q, E and C

increase

3. Q remains unchanged, C increases, V and

E decreases

4. Q & C increase but V & E decrease

Two condensers, one of capacity C and the other of capacity $\frac{\mathrm{C}}{2}$ , are connected to a V volt battery, as shown

1. $2{\mathrm{CV}}^{2}$

2. $\frac{1}{4}C{V}^{2}$

3. $\frac{3}{4}C{V}^{2}$

4. $\frac{1}{2}C{V}^{2}$

The energy required to charge a parallel plate

condenser of plate separation d and plate area of

cross-section A such that the uniform electric field

between the plates is E, is

1. ${\mathrm{\epsilon}}_{0}{\mathrm{E}}^{2}/\mathrm{Ad}$

2. ${\mathrm{\epsilon}}_{0}{\mathrm{E}}^{2}\mathrm{d}$

3. $\frac{1}{2}{\epsilon}_{0}{E}^{2}Ad$

4. $\frac{1}{2}{\epsilon}_{0}{E}^{2}\mathit{/}Ad$

In the figure below, the capacitance of each capacitor

is 3$\mathrm{\mu}$F. The effective capacitance between A and B is

1. $\frac{3}{4}\mu F$

2. $3\mathrm{\mu F}$

3. $6\mathrm{\mu F}$

4. $5\mathrm{\mu F}$

The equivalent capacity between the points X and Y

in the circuit with C = 1 $\mathrm{\mu}$F is

1. $2\mathrm{\mu F}$

2. $3\mathrm{\mu F}$

3. $1\mathrm{\mu F}$

4. $0.5\mathrm{\mu F}$

A parallel plate capacitor with air as dielectric is

charged to a potential ‘V’ using a battery. Removing

the battery, the charged capacitor is then connectged

across an identical uncharged parallel plate capacitor

filled with wax of dielectric constant ‘k’. The common

potential of both the capacitors is

1. V volts

2. kV volts

3. (k+1) V volts

4. $\frac{\mathrm{V}}{\mathrm{k}+1}volts$

What is the potential difference across 3 $\mathrm{\mu}$F capacitor

in the circuit shown in the figure?

1. 6V

2. 2V

3. 4V

4. 16V

A parallel plate capacitor with air as the dielectric has

capacitance C. A slab of dielectric constant K and

having the same thickness as the separation between

the plates is introduced so as to fill one-fourth of the

capacitor as shown in the figure. The new capacitance

will be

1. $\left(\mathrm{K}+3\right)\frac{C}{4}$

2. $\left(\mathrm{K}+2\right)\frac{C}{4}$

3. $\left(\mathrm{K}+1\right)\frac{C}{4}$

4. $\frac{\mathrm{KC}}{4}$

In the given circuit diagram, initially battery was

connected. Find the work done by battery if capacitor

is completely filled with a dielectric of dielectric

constant k = 3.

1. $\frac{1}{2}C{V}^{2}$

2. ${\mathrm{CV}}^{2}$

3. $2{\mathrm{CV}}^{2}$

4. $\frac{3}{2}C{V}^{2}$

The electric potential V at any point x, y, z (all in

metres) in space is given by V = 4x^{2} volts. The electric

field (in V/m) at the point (1 m, 0, 2 m)

1. $-8\hat{\mathrm{i}}$

2. $8\hat{\mathrm{i}}$

3. $-16\hat{\mathrm{i}}$

4. $8\sqrt{5}\hat{\mathrm{i}}$

In a parallel-plate capacitor of capacitance C, a metal

sheet is inserted between the plates, parallel to them.

The thickness of the sheet is half of the separation

between the plate. The capacitance now becomes

1. 4C

2. 2C

3. C/2

4. C/4

Capacitance of a capacitor becomes $\frac{4}{3}$ times its

original value if a dielectric slab of thickness t = d/2 is

inserted between the plates (d = separation between

the plates). The dielectric constant of the slab is

1. 2

2. 4

3. 6

4. 8

In the figure shown, conducting shells A and B have

charges Q and 2Q distributed uniformly over A and

B .Value of V_{A} – V_{B} is

1. $\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon}}_{0}\mathrm{R}}$

2. $\frac{\mathrm{Q}}{8{\mathrm{\pi \epsilon}}_{0}\mathrm{R}}$

3. $\frac{3\mathrm{Q}}{4{\mathrm{\pi \epsilon}}_{0}\mathrm{R}}$

4. $\frac{3\mathrm{Q}}{8{\mathrm{\pi \epsilon}}_{0}\mathrm{R}}$

Three point charges q, –2q and –2q are placed at the

vertices of an equilateral triangle of side a. The work

done by some external force to slowly increase their

separation to 2a will be

1. $\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\frac{2{q}^{2}}{a}$

2. $\frac{{q}^{2}}{4{\mathrm{\pi \epsilon}}_{0}\mathrm{a}}$

3. $\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\frac{\mathit{3}{q}^{2}}{3R}$

4. Zero

Two capacitors of capacitance 3 $\mathrm{\mu}$ F and 6 $\mathrm{\mu}$ F are

charged to a potential of 12 V each. They are now

connected to each other, with the positive plate of

one to the negative plate of the other. Then

1. the potential difference across 3 $\mathrm{\mu}$ F is zero

2. the potential difference across 3 $\mathrm{\mu}$ F is 4 V

3. the charge on 3 $\mathrm{\mu}$ F is zero

4. the charge on 3 $\mathrm{\mu}$ F is 10 $\mathrm{\mu}$ C

Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that (E = electric field, V = potential)

1. V_{A} = V_{B}

2. E_{A} = E_{B}

3. V_{A} V_{B}

4. V_{A} < V_{B}

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