Physics, asked by shivani5775, 11 months ago

electric field strength due to a point charge of 5 microcoulomb at a distance of 50 cm from the charge is​

Answers

Answered by rahulchugh33333
1

Answer:

Explanation:

E=KQ/R2

E=9*109 *5*10-6/.5*.5

E=1.8*105 N/C

Similar questions