Question

a point charge of 0.33×10^-8 coulomb is placed in a medium of relative permitivity of 5 calculate electric field intensity at a point 10 cm from the charge ​

Answer 1

Answer:

5. 94×10² N/C

Explanation:

=9×10^9/5×0.33×10^¯8/(1/10)²

=0.594×10³N/C

=5.94×10²N/C