Question

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

Answer 1

Answer:

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Answer 2

Given :

The magnitude of charge 1 = $Q_1$ = 3 × $10^{-5}$ C

The magnitude of charge 2 = $Q_2$ = 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 × $\dfrac{Q_1Q_2}{(distance )^{2} }$

or    F =  k × $\dfrac{Q_1Q_2}{(d )^{2} }$              ,  where k = 9 × $10^{9}$

i.e    F = 9 × $10^{9}$ × $\dfrac{3\times 10^{-5} \times 5\times 10^{4}}{(0.1 m)^{2}}$

Or, F =  9 × $10^{9}$ ×  $\dfrac{1.5}{0.01m^{2}}$

∴   F = 9 × $10^{9}$ × 150

i.e F = 1350 × $10^{9}$

or,  Force = F = 13.5 × $10^{11}$ N

Hence, The magnitude of electrostatic force acting between charges is  13.5 × $10^{11}$ Newton  . Answer