Name: Equipotential Surfaces and Its Properties
Uploaded: 2019-07-19
Duration: 7 min 22 s
Description: The surface over which the electric potential is the same everywhere is called an equipotential surface. Let's more about it and its properties in this video.

Question

Two tiny metal spheres $A$ and $B$ , mass $m_{A} = 5.00 g$ and $m_{B} = 10.0 g$ , have equal positive charge $q = 5.00 μ C .$ The spheres are connected by a massless nonconducting string of length $d = 3.00 m,$ which is much greater than the radii of the spheres.

(a) What is the electric potential energy of the system?
(b) Suppose you cut the string. At that instant, what is the acceleration of each sphere?

(c) A long time after you cut the string, what is the speed of each sphere?

Accepted Answer

(a) The potential energy is

$B$ ,

relative to the potential energy at infinite separation.

(b) Each sphere repels the other with a force that has magnitude

$F = \frac{q^{2}}{4 π ε_{0} d^{2}} = \frac{(9 \times 10^{9} N m^{2} C^{- 2}) {(5.0 \times 10^{- 6} C)}^{2}}{(3.00 m)^{2}} = 0.025 N$ .

According to Newton's second law the acceleration of each sphere is the force divided by the mass of the sphere. Let $m_{A}$ and $m_{B}$ be the masses of the spheres. The acceleration of sphere $A$ is

$a_{A} = \frac{F}{m_{A}} = \frac{0.025 N}{5.0 \times 10^{- 3} kg} = 5.0 m s^{- 2}$

and the acceleration of sphere $B$ is

$a_{B} = \frac{F}{m_{B}} = \frac{0.025 N}{10 \times 10^{- 3} kg} = 2.5 m s^{- 2}$ .

(c) Energy is conserved. The initial potential energy is $U = 0.075 J$ , as calculated in part (a). The initial kinetic energy is zero since the spheres start from rest. The final potential energy is zero since the spheres are then far apart.

The final kinetic energy is $\frac{1}{2} m_{A} v_{A}^{2} + \frac{1}{2} m_{B} v_{B}^{2}$ ,

where, $v_{A}$ and $v_{B}$ are the final velocities. Thus,

$U = \frac{1}{2} m_{A} v_{A}^{2} + \frac{1}{2} m_{B} v_{B}^{2}$ .

Momentum is also conserved, so

$0 = m_{A} v_{A} + m_{B} v_{B}$ .

These equations may be solved simultaneously for $v_{A}$ and $v_{B}$ . Substituting $v_{B} = - (\frac{m_{A}}{m_{B}}) v_{A}$ from the momentum equation into the energy equation, and collecting terms, we obtain

$U = \frac{1}{2} m_{A} {v^{2}}_{A} + \frac{1}{2} m_{B} {(\frac{m_{A}}{m_{B}})}^{2} {v^{2}}_{A}$ .

Thus,

$v_{A} = \sqrt{\frac{2 U m_{B}}{m_{A} (m_{A} + m_{B})}} = \sqrt{\frac{2 (0.075 J) (10 \times 10^{- 3} kg)}{(5.0 \times 10^{- 3} kg) (5.0 \times 10^{- 3} kg + 10 \times 10^{- 3} kg)}} = 4.47 m s^{- 1}$ .

We thus obtain

$v_{B} = - \frac{m_{A}}{m_{B}} v_{A} = - (\frac{5.0 \times 10^{- 3} kg}{10 \times 10^{- 3} kg}) (4.47 m s^{- 1}) = - 2.23 m s^{- 1}$

or $|v_{B}| = 2.23 m s^{- 1}$ .

Important Questions on Electric Potential

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