Question

When a particle carrying a charge of 10 into 10 to the power minus 6 coulomb is brought from infinity to a point speed 2 into 10 to the power minus 3 joule of work is done what is the potential at the point p?

Answer 1

Given,

when a particle carrying a charge of 10 × 10^-6 C is brought from infinity to a point P where 2 × 10¯³ J of work is done.

To find,

potential at point P.

magnitude of charge of the particle, Q = 10 × 10^-6 C = 10^-5 C

work done on the particle, W = 2 × 10¯³ J

now potential at point P = W/Q

= 2 × 10¯³/10^-5

= 200 volts

Therefore potential at point P is 200 volts.

Answer 2

Answer:

magnitude of charge of the particle, Q = 10 × 10^-6 C = 10^-5 C

work done on the particle, W = 2 × 10¯³ J

now potential at point P = W/Q

= 2 × 10¯³/10^-5

= 200 volts