Question

Calulate the force experienced by test charge of 0.2µc placed in an electric field of 4.2×106 N/C

Answer 1

Answer:

The electric field E is defined to be E=Fq E = F q , where F is the Coulomb or electrostatic force exerted on a small positive test charge q. ... The magnitude of the electric field E created by a point charge Q is E=k|Q|r2 E = k | Q | r 2 , where r is the distance from Q.

Explanation:

SOLVE SOLVE WITH THIS SOLUTION AND THIS FORMULA

Answer 2

Answer:

0.84 N

Explanation:

F = qe

[0.2*(10)-6]* [4.2*( 10)6]

0.2*4.2

0.84N