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Kipish [7]
2 years ago
13

Two positive point charges, each with charge q, separated by a distance d, repel each other with a force of magnitude 20 N. What

is the magnitude of the force between two positive point charges of magnitude 11.11 q, separated by a distance 2.5 d in units of N
Physics
1 answer:
Doss [256]2 years ago
3 0

Answer:

The value is  F_2 =  395 \  N

Explanation:

From the question we are told that

   The magnitude of the charge of each positive charge for the first case is q_1 = q_2 =  q

   The distance between the charges for the first case is  d

   The  force between the charges for the first case  is  F =  20 \ N

     The magnitude of the charge of each positive charge for the second case is q_1 = q_2=11.11q

     The  distance between the charge for the second case is 2.5d

 

Generally for the first case the force between the charge is mathematically  represented as

       F_1 = \frac{k * q^2 }{d^2}

Where k is the Coulomb constant with value   k =  9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

  So  

     20= \frac{k * q^2 }{d^2}

Generally for the second  case the force between the charge is mathematically  represented as

       F_2 =  \frac{k *  (11.11q)^2}{(2.5d)^2}

       F_2 =  \frac{k * 11.11^2 *q^2}{2.5^2d^2}

=>      F_2 =  \frac{11.11^2}{2.5^2}  F_1

=>      F_2 =  \frac{11.11^2}{2.5^2} * 20

=>      F_2 =  395 \  N

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