You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

The Self Potential Energy of Continuous Charge Distributions

You are movin...

Question

You are moving a negative charge -q, where -q<0 at a small constant speed away from a uniformly charged non-conducting spherical shell on which resides a negative charge -Q<0. The electrostatic field of -Q is E. Let U be the total energy of the system, $W_{a}$ is the work done by the force $F_{a}$ you exert on -q and $W_{E}$ the work done by the electrostatic force $F_{E}$ on -q. Then, as -q is being moved

W_{a} = - W_{E}

, therefore U remains constant

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F_{a} = - F_{E}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

U increases

No worries! We‘ve got your back. Try BYJU‘S free classes today!

U decreases

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D U decreases

If charge is being moved slowly, $_{a} = -_{E}$
$W_{a} + W_{E} = Δ K E = 0$ [Charge is moved slowly at constant speed]
i.e., $W_{a} = - W_{E}$
Also $W_{E} = - Δ U$
Thus, $W_{a} = + Δ U$
As force $F_{a}$ and displacement are $180^{\circ}$
$\to F . \to d r < 0$
$W_{a} < 0$
Thus, $Δ U < 0$
therefore, U decreases.

Suggest Corrections

Similar questions

Q. You are moving a negative charge

q < 0

at a small constant speed away from a uniformly charged non-conducting spherical shell on which resides a negative charge

Q < 0

. The electrostatic field of

Q

E

. Let

U

be the total energy of the system,

W_{a}

the work done by the force

F_{a}

you exert on

q

and

W_{E}

the work done by the electrostatic force

F_{E}

q

. Then, as

q

is being moved,

Q. The lines of force of the electric field of a positive charge

(+ q)

and a negative charge

(- q)

are shown in figure (a) and (b) below. Then:-

(a) The work done in moving a small positive charge (+

q_{0}

) from Q to P will be positive
(b) The work done in moving a small negative charge (−

q_{0}

) from B to A will be positive
(c) In going from Q to P, the kinetic energy of a small negative charge (−

q_{0}

) increases
(d) In going from B to A, the kinetic energy of a small negative charge (−

q_{0}

) decreases

Work done by electrostatic force to take point charge

Q

from O to O' is. {Where O and O' are center of uniformly charged rings of radius

R

with charge

+ Q

and

Q

. If

\frac{k Q^{2}}{R} = 1

}

Q. A spherical shell with an inner radius 'a' and an outer radius 'b' is made of conducting material. A point charge +Q is placed at the center of the spherical shell and a total charge -q is placed on the shell.
(a) Charge

- q

is distributed on the surfaces as
(A)

- Q

on the inner surface,

- Q

on outer surface
(B)

- Q

on the inner surface,

- q + Q

on outer surface
(C)

+ Q

on the inner surface,

- q - Q

on outer surface
(D) The charge

- q

is spread uniformly between the inner and outer surface.
(b) Assume that the electrostatic potential is zero at an infinite distance from the spherical shell. The electrostatic potential at a distance

R (a < R < b)

from the centre of the shell is

[w h e r e K = \frac{1}{4 π ε_{0}}]

A positive point charge +Q is fixed in space. A negative point charge -q of mass m revolves around a fixed charge in elliptical orbit. The fixed charge +Q is at one focus of the ellipse. The only force acting on negative charge is the electrostatic force due to positive charge. Then which of the following satement is true.

Join BYJU'S Learning Program

The correct option is D U decreases If charge is being moved slowly, −→Fa=−−→FE Wa+WE=ΔKE=0 [Charge is moved slowly at constant speed] i.e., Wa=−WE Also WE=−ΔU Thus, Wa=+ΔU As force Fa and displacement are 180∘ →F.→dr<0 Wa<0 Thus, ΔU<0 therefore, U decreases.