Two charges 3*10^-5 C and 5*10^4 C are placed at a distance 10 cm form each other. Find the value of electrostatic force acting between them. a) 13.5 x 10^11N b) 40 x 10^11N c)180 x 10^9N d)13.5 x 10^10 N

13.5 x 10^(11) N

Explanation:

Given :

The magnitude of charge 1 = q1 = 3 × 10^(-5) C

The magnitude of charge 2 = q2 = 5 × 10^(4) C

The distance between charges = d = 10 cm = 0.1 m

To Find :

The magnitude of electrostatic force acting between charges

Solution :

∵ Electrostatic force = F = (k × (q1) x (q2)) / r^(2)

where k = 9 × 10^(9)

i.e    F = 9 × 10^(9) × (3 × 10^(-5) × 5 × 10^(4)) / (0.1)^(2)  

Or, F =  9 × 10^(9) ×  (1.5/0.01)  

∴   F = 9 × 10^(9) × 150

i.e F = 1350 × 10^(9)

or,  Force = F = 1.35 × 10^(12) N

Hence, The magnitude of electrostatic force acting between charges is  1.35 × 10^(12) N => 13.5 x 10^(11)

+15-35=-20

going in opposite directions

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