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Electric field is given by E $= \frac{100}{x^{2}}$ . Find the potential difference between x= 10 and x= 20 m. [This question is only for Dropper and XII batch]

5 V

10 V

15 V

20 V

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Explanation

$E = - \frac{d V}{d x} \Rightarrow d V = - E d x \Rightarrow {}_{A}^{B}\int d V = - {}_{A}^{B}\int E . d x \Rightarrow V_{B} - V_{A} = - {}_{10}^{20}\int \frac{100}{x^{2}} = - 5 v o l t s P o t e n t i a l d i f f e r e n c e = 5 v o l t .$

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Calculate the electric field on the axis of a very long uniformly charged, thin rod at a distance r from one end. The charge per unit length of the rod is $λ$ . [This question is only for Dropper and XII batch]

$\frac{2 k λ}{r}$

$\frac{k λ}{r}$

$\frac{k λ}{2 r}$

$\frac{k λ}{4 r}$

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Find electric field at the centre of arc which subtends ' $θ$ ' angle at centre

[This question is only for Dropper and XII batch]

$\frac{2 k λ}{r} \sin θ$

$\frac{2 k λ}{r} \sin \frac{θ}{4}$

$\frac{2 k λ}{r} \sin \frac{θ}{2}$

$\frac{2 k λ}{r} \sin 2 θ$

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Explanation

The electric field at the center of a uniformly charged arc subtending an angle Î¸ at the center is given by (2kÎ»/r)sin(Î¸/2), where k is the Coulomb constant, Î» is the charge per unit length, and r is the radius of the arc. This expression takes into account the symmetry of the charged arc.

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Find electric field due to a uniformly charged, long and thin rod

[This question is only for Dropper and XII batch]

$\frac{k λ}{r}$

$\frac{k λ}{2 r}$

$\frac{2 k λ}{r}$

$\frac{k λ}{4 r}$

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A ball of radius R carries a positive charge throughout its volume, such that the volume density of charge depends on distance r from the ball's centre as $ρ = ρ_{0} (1 - \frac{r}{R})$ , where $ρ_{0}$ is a constant. Assuming the permittivity of ball to be one, find magnitude of electric field as a function of distance r, both inside the ball. [This question is only for Dropper and XII batch]

$\frac{ρ_{0} r}{3 ε_{0}}$

$\frac{ρ_{0} (4 r R - 3 r^{2})}{12 ε_{0} R}$

$\frac{ρ_{0} (4 r R - 3 r^{2})}{3 ε_{0} R}$

$\frac{ρ_{0} (4 r R - 5 r^{2})}{3 ε_{0} R}$

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The electric field vector in a region given by $\vec{E} = (3 \hat{i} + 4 y \hat{j}) V m^{- 1}$ . Calculate the potential at (1m, 1m) taking potential at origin to be zero. [This question is only for Dropper and XII batch]

-1V

-5V

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Explanation

Here, $\vec{E} = \hat{i} + 4 y \hat{j}, \vec{r} = 0 \hat{i} + 0 \hat{j}, \vec{r_{2}} = \hat{i} + \hat{j}, d r = d x \hat{i} + d y \hat{j}, V_{1} = 0, V_{2} = V$

$\Rightarrow V - 0 = - \int_{T 1}^{T 2} (3 \hat{i} + 4 y \hat{j}) . (d x \hat{i} + d y \hat{j}) \Rightarrow V - 0 = - {}_{0}^{1}\int 3 d x - {}_{0}^{1}\int 4 y d y \Rightarrow V = - 3 - 4 {[\frac{y^{2}}{2}]}_{0}^{1} = - 3 - \frac{4}{2} = - 5 V$

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Each of a parallel-plate air capacitor has an area S. What amount of work has to be performed to slowly increase the distance between the plates from $x_{1} t o x_{2}$ . If the voltage across the capacitor, which is equal to V, is kept constant in the process. [This question is only for Dropper and XII batch]

$\frac{ε_{0} S V}{2}$

$ε_{0} S V^{2} [\frac{1}{x_{1}} - \frac{1}{x_{2}}]$

$\frac{ε_{0} S V^{2}}{2} [\frac{1}{x_{1}} - \frac{1}{x_{2}}]$

$ε_{0} S V$

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Explanation

When voltage is kept constant, the force acting on each plate of capacitor will depend on the distance between the plates.

So, elementary work done by agent, in its displacement over a distance dx, relative to the other,
dW= $- F_{x} d x$

But, $F_{x} = - (\frac{σ (x)}{2 ε_{0}}) S σ (x) a n d σ (x) = \frac{ε_{0} V}{x}$

Hence, $W = \int d W = {}_{x 1}^{x 2}\int \frac{1}{2} ε_{0} \frac{S V^{2}}{x^{2}} d x = \frac{ε_{0} S V^{2}}{2} [\frac{1}{x_{1}} - \frac{1}{x_{2}}]$

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Using the concept of energy density, find the total energy stored by a shell of radius R and charge Q. [This question is only for Dropper and XII batch]

$\frac{Q^{2}}{{πε}_{0} R}$

$\frac{Q^{2}}{2 {πε}_{0} R}$

$\frac{Q^{2}}{4 {πε}_{0} R}$

$\frac{Q^{2}}{8 {πε}_{0} R}$

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A ball is thrown upward with a certain speed. It passes through the same point at 3 second and 7 second from the start. The maximum height achieved by the ball is

500 m

250 m

125 m

450 m

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Six particles situated at the corners of a regular hexagon of side a move at constant speed v. Each particle maintains a direction towards the particle at the next. The time which the particle will take to meet each other is

$\frac{2 a}{v} \sec$

$\frac{a}{v} \sec$

$\frac{2 a}{3 v} \sec$

$\frac{3 a}{v} \sec$

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