Question

A particle with a charge of +4.0 μC has a mass of 5.0 g. What magnitude electric field directed upward will exactly balance the weight of the particle?

Answer 1

Answer:

E = 12262 V/m

Explanation:

given,

Charge of the particle, q = +4.0 μC

mass of the ball, m = 5 g

Electric field, E = ?

Force due to weight

F = m g

Force due to electric field

F =q E

To balance the weight of particle both the forces must be equal

Electric field exerted on the ball will be equal to

m g = q E

$E = \dfrac{m\ g}{q}$

$E = \dfrac{5\times 10^{-3}\times 9.81}{4\times 10^{-6}}$

E = 12262 V/m

Hence, the electric field acting in the particle is equal to E = 12262 V/m