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Electrostatic Potential and Capacitance

Multiple Choice Questions

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The plates of a parallel plate capacitor of capacity of 50 µF are charged by a battery to a potential of 100 volt. The battery remains connected the plates are separated from each other so that the distance between them is doubled. Then, the energy spent by battery in doing so, will be

$12.5 \times 10^{- 2} J$
$35 \times 10^{- 2} J$
42.5×10^-2J
25×10^-2J

25×10^-2J

When the separation between the plates is doubled the capacitance becomes one half.

It means C'=50/2=25μF

Now energy spent by the battery =qV=(C'V)V

=C'V²=25×10^-6×(100)²

=25×10^-2J

From the figure find the capacitance of the capacitor?

The given capacitor can be viewed as a series combination of two capacitors C₁ and C₂ where

where d is the seperation between the plates

∴ The effective capacitance

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2µF. The separation is reduced to half and it is filled with a dielectric substance of value of 2.8. The final capacity of capacitor is

15.6μF
11.2μF
19μF
22.4μF

11.2μF

$C = \frac{ε_{0} A}{d} \propto \frac{1}{d} Hence, \frac{C_{1}}{C_{2}} = \frac{d_{2}}{d_{1}} or \frac{2}{C_{2}} = \frac{0.2}{0.4} or C_{2} = \frac{0.4 \times 2}{0.2} = 4 μF Final capacitor is C = 4 \times k = 4 \times 2.8 = 11.2 μF$

If a unit positive charge is taken from one point to another over an equipotential surface, then

work is done on the charge
work is done by the charge
work done is constant
no work is done

no work is done

Since the potential at each point of an equipotential surface is the same. the potential does not change while we move a unit positive charge from one point to another. Therefore work done in the process is zero.

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Two small magnets each of magnetic moment 10 Am² are placed in end on position 0.1 m apart from their centres. The force acting between them is

0.6N
0.06N
0.06×10⁷N
0.6×10⁷N

0.6N

$F = \frac{μ_{o}}{4 π} \frac{6 M_{1} M_{2}}{r^{4}} = 10^{- 7} \times \frac{6 \times 10 \times 10}{{(0.1)}^{4}} = 0.6 N$

An electron is travelling along the x-direction. It encounters a magnetic field in the y-direction. Its subsequent motion will be

straight line along the x-direction
a circle in the xz plane
a circle in the yz plane
a circle in the xy plane

a circle in the xz plane

We know that the force on a charged particle e due to a magnetic field is given by

$\vec{F} = e (\vec{v} \times \vec{B})$

Here $\vec{v} = v_{x} = \hat{i} and \vec{B} = B_{y} \hat{j}$

∴ $\vec{F} = e v_{x} B_{y} (\vec{i} \times \vec{j})$

$\vec{F} = {ev}_{x} B_{y} \hat{k}$

Therefore the subsequent motion of charged particle will be a circle in the xz plane.

Two metal plates having a potential difference of 800 V and 2 cm apart. It is found that a particle of mass 1.96 x 10^-15 kg remaining suspended in the region between the plates. The charge on the particle must be (e = elementry charge)

$E = \frac{V}{d} = \frac{800}{2 \times 10^{- 2}} \frac{V}{m = 4 \times 10^{4} \frac{V}{m}} again, qE = mg q = \frac{mg}{E} = \frac{1.96 \times 10^{- 15} \times 9.8}{4 \times 10^{4}} e q = 3 e$

An electron having charge 'e' and mass m is moving in a uniform electric field E. Its acceleration will be

$\frac{e^{2}}{m}$
$\frac{eE}{m}$
$\frac{{eE}^{2}}{m}$
$\frac{mE}{e}$

$\frac{eE}{m}$

Charge on electron = e

Mass of electron = m

intensity of uniform electric field = E

Acceleration of the electron

$α = \frac{Force on the electron}{Mass of electron}$

= $\frac{Electric charge \times electric field}{Mass of electron}$

$α = \frac{eE}{m}$

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In the given figure, the capacitance C₁, C₃, C₄ and C₅ have a capacitance 4 μF each. If the capacitor C₂ has a capacitance 10 μF, then effective capacitance between A and B will be

2 μF
6 μF
4 μF
8 μF

4 μF

Capacitance of capacitors C₁ , C₃ , C₄ , C₅ = 4 μF each and capacitance of capacitor C₂ = 10 μF.

If a battery is applied across A and B, the points b and c will be at the same potential ( since C₁ = C₄ = C₃ = C₅ = 4 μF )

Therefore no charge flows through C₂.

We have the capacitors C₁ and C₅ in series.

Therefore their equivalent capacitance

C'' = $\frac{C_{3} \times C_{5}}{C_{1} + C_{5}}$

= $\frac{4 \times 4}{4 + 4}$

C" = 2 μF

Now C' and C" are in parallel. Therefore capacitance between A and B

= C' + C"

= 2 + 2

= 4 μF

10.

The working of dynamic is based on the principle of

Electromagnetic induction
chemical effect of current
magnetic effect of current
heating effect of current

Electromagnetic induction

The dynamo operates on the principle of the production of dynamically induced emf. Hence whenever flux is cut by the conductor, emf is produced in it according to the law of electromagnetic induction.

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