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)

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .