5. A 5.5 x10-6 C charge is located 0.28 m from a -3.5 x 10-6 C charge.
a) What is the magnitude of the force that the positive charge exerts on the negative charge?
b) If the negative charge was doubled, how would the force change?

(a) The magnitude of force that the positive charge exerts on the negative charge is - 2.209 N.

(b) If the negative charge is doubled, then the force will also get doubled.

The new force will be F = -4.418 N.

Explanation:

The force acting between two charged particles separated by a distance is termed as Coloumb's force or electrostatic force. It can be termed as electrostatic force of attraction if the the force acting between the charges are oppositely charged. And it can be termed as electrostatic force of repulsion if the charges are similar or like charges.

In the present case, there is a positive and negative charge, so electrostatic force of attraction will be acting between them. As per Coloumb's law, the electrostatic force of attraction is directly proportional to the product of charges and inversely proportional to the square of distance of separation.

Here, k is the constant of proportionality which is equal to 9 × and Q, q are the two charges, d is the distance of separation.

So here Q = 5.5 × and q = - 3.5 × and d = 0.28 m

Then,

So the magnitude of force that the positive charge exerts on the negative charge is - 2.209 N.

(b) If the negative charge is doubled, then the force will also get doubled.

The new force will be F = -4.418 N.

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