Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are doubled and their separation is halved, what happens to the electrical force that each charge exerts on the other one

F' = 16 F

Hence, the electric force between charges becomes sixteen times its initial value.

Explanation:

The electric force between the two charges is given by the Colomb's Law:

------------------- eq(1)

where

F = electric force

K = Colomb's Constant

Q₁ = magnitude of the first charge

Q₂ = magnitude of the second charge

R = Distance between charges

Now the magnitudes of the charges are doubled and the distance between them is halved. Therefore:

using equation (1):

F' = 16 F

Hence, the electric force between charges becomes sixteen times of its initial value.

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27