﻿ IIT - JEE Main Important Questions of Electrostatic Potential and Capacitance | Zigya

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# Electrostatic Potential and Capacitance

#### Multiple Choice Questions

1.

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 µF and 9 µF capacitors), at a point distant 30 m from it, would equal:

• 240N/C

• 360N/C

• 420N/C

• 480N/C

C.

420N/C

Resultant circuit,

As, charge on 3μF = 3μF x 8V = 24μC
Charge on 3μF = 3μF x 2V = 18 μC
charge on 4μF +Charge on 9μF
= (24 + 18)μC = 42μC
therefore,
Electric field at a point distant 30 m

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2.

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to

• 6 x 10-7 C/m2

• 3  x 10-7 C/m2

• 3 x 104 C/m2

• 6x 104 C/m2

A.

6 x 10-7 C/m2

When free space between parallel plate capacitor,
When dielectric is introduced between parallel plates of capacitor,

Electric field inside dielectric

where, K = dielectric constant of medium = -2.2
εo = permitivity of free space = 8.85 x 10-12
σ = 2.2 x 8.85 x 10-12 x 3 x 104
= 6.6 x 8.85 x 10-8 = 5.841 x10-7
= 6 x 10-7 C/m2

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3.

The electrostatic potential inside a charged spherical ball is given by Φ = ar2 + b where r is the distance from the centre; a,b are constants. Then the charge density inside the ball is

• -6aε0r

• -24πaε0

• -6aε0

• -24πaε0r

C.

-6aε0

Electric field,

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4.

A fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is

A.

As initially, charge is maximum
∴ q = qocos ωt
Current,

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5.

Assume that an electric field  exists in space. Then the potential difference VA – VO, where VO is the potential at the origin and VA the potential at x = 2 m is:

• 120 J

• -120 J

• -80 J

• 80 J

C.

-80 J

As we know, potential difference VA -Vo is
dV = - Edx

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6.

A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ) on its surface. For this sphere, the equipotential surfaces with potentials 3Vo/2, 5Vo/4, 3V/4 and Vo/4 have radius R1, R2,R3 and R4 respectively. Then

• R1 = 0 and R2>(R4-R3)

• R1≠0 and (R2-R1)>(R4-R3)

• R1 = 0 and R2<(R4-R3)

• 2R<R4

C.

R1 = 0 and R2<(R4-R3)

The potential at the surface of the charged sphere.

As potential decreases for outside points. Thus, according to the question, we can write

Similarly,

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7.

The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant τ of this circuit lies between

• 150 sec and 200 sec

• 0 and 50 sec

• 50 sec and 100 sec

• 100 sec and 150 sec

D.

100 sec and 150 sec

Time constant τ is the duration when the value of potential drops by 63% of its initial maximum value (i.e, Vo/e)
Here, 37% of 25 V = 9.25 V which lies between 100s to 150 s in the graphs.

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8.

Charge +q and -q are placed at points A and B respectively which are distance 2L apart, C is the midpoint between A and B.The work done in moving a charge + Q along the semicircle CED is:

D.

work done is equal to change in potential energy.

In Ist case, when charge + Q is situated at C.

Electric potential energy of system

In IInd case, when charge + Q is moved from C to D.

Electric potential energy of system in that case

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9.

In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF.  Q2 as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale)

A.

From the given circuit,
Qc= Q1+Q2
or
C (E-V) = 1 x V + 2 XV
Or V (C+3) = CE

As, C1 varied from 1μF to 3μF, charge increases with decreasing slope.

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10.

Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then

• 5C1 = 3C2

• 3C1 = 5C2

• 3C1 = 5C2 = 0

• 9C1 = 4C2

B.

3C1 = 5C2

q1 = q2
C1V1 = C2V2
120C1 = 200C2
⇒ 3C1 = 5C2

418 Views