Question

Two charges repel each other with a force of 200N. One of the charges is increased by 20% and the other is decreased by 10% . The new repulsive force at the same distance would be

Answer 1

NEW FORCE IS 216 NEWTON.

Given:

• Initial force is 200 N
• One charge is increased by 20% and another is decreased by 10%.

To find:

New repulsive forces at same distance?

Calculation:

Let the charges be Q and q.

So, initial force is :

$f = \dfrac{kQq}{ {r}^{2} } = 200N$

Now, new charges are :

• Q' = Q × 120/100 = 1.2Q
• q' = q × 90/100 = 0.9q

So, new force is :

$f_{new} = \dfrac{k(1.2Q)(0.9q)}{ {r}^{2} }$

$\implies f_{new} = 1.08 \times \dfrac{kQq}{ {r}^{2} }$

$\implies f_{new} = 1.08 \times 200$

$\implies f_{new} = 216 \: N$

So, new force is 216 N.

Answer 2

Answer:

New force is : 216 N

Explanation:

Force = 1 / 4 π∈0 * q1 *q2/ r^2 = 200N

q1 ' = q1 *(1+ 20/ 100) =1.2 q1

q2 ' =q2* (1- 10 / 100)= 0.9q2

new charge increase is  q1 ' = 1.2 q1

new charge decrease is q2 ' = 0.9q2

| F ' | = 1 / 4 π∈0 * ( 1.2 Q1) ( 0.9 Q2) / R^2

|F ' | = 1.2 * 0.9 * 200 = 216N        

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