Question

a particle of mass 10 mg is in equilibrium in a verticle electric field E=10 raise to power 5 j cap the charge on particle is (g= -10j cap)​

Answer 1

mass of particle, m = 10mg = 10^-6 Kg

electric field intensity, $\vec{E}=10^5\hat{j}$

Let the charge on the particle is q.

so, force acting due to electric field on charged particle , F = qE

= q × $10^5\hat{j}$ N

gravitational force acting on body, W = mg

= $10^{-6}Kg\times(-10\hat{j})$

= $-10^{-5}\hat{j}$ N

net force acting on particle , Fnet = F + W

= $q 10^5\hat{j}-10^{-5}\hat{j}$

at equilibrium,

Fnet = 0

so, q × 10^5 j = 10^-5 j

or we can write , 10^5 q = 10^-5

so, q = 10^-10 C

hence, charge on particle is 10^-10 C

Answer 2

Answer:

Hi! I have explained the answer in the photo I have attached

Explanation:

P.S. The other person has not explanaed the question correctly, since they did not take into account the correct value of mass.