Question

two point charges repel each other with a force of 100N .if one charge is increased by 10% and other is reduced by 10% .the new force of repulsion at the same distance would be..?​

Answer 1

Answer:

New force of repulsion = 99 N

Explanation:

Given:

• Two point charges repel each other with a force of 100 N
• One charge is increased by 10% and the other charge is reduced by 10%

To Find:

The new force of repulsion

Solution:

By Coulomb's law we know that the force between two point charges is given by,

$\boxed{\sf F =k\times \frac{q_1q_2}{r^2}}$

where k is the Coulomb's law constant, F is the force, q₁, q₂ are the point charges and r is the distance between the two charges.

The initial force of repulsion is given by,

$\sf F_1=k\times \dfrac{q_1q_2}{r^2} =100\:N$

Now given that the first charge is increased by 10%

Hence,

(q₁)' = q₁ + q₁ × 10/100

(q₁)' = q₁ + q₁/10

(q₁)' = 11 q₁/10-------(1)

Also by given second charge is decreased by 10%

Hence,

(q₂)' = q₂ - q₂ × 10/100

(q₂)' = 9 q₂/10 ------(2)

Now the new force of repulsion is given by,

$\sf F_2=k\times \dfrac{(q_1)'(q_2)'}{r^2}$

Substitute the values from 1 and 2,

$\sf F_2=k\times \dfrac{(11\:q_1/10)(9\:q_2/10)}{r^2}$

$\sf F_2=\dfrac{99}{100}\times k\times \dfrac{q_1q_2}{r^2}$

Hence,

F₂ = 0.99 × F₁

F₂ = 0.99 × 100

F₂ = 99 N

Hence the new force of repulsion is 99 N.

Answer 2

Answer:

⠀⠀⠀⌬ Force = 100N

⠀⠀⠀⌬ One Charge (a) = + 10%

⠀⠀⠀⌬ Other Charge (b) = - 10%

⠀⠀⠀⌬ New Force = ?

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf New\:Force=Force+Relative\:Charge\\\\\\:\implies\sf New\:Force=Force+\bigg\lgroup a+b+\dfrac{ab}{100}\bigg\rgroup\\\\\\:\implies\sf New\:Force=100+\bigg\lgroup 10+( - 10)+\dfrac{10 \times ( - 10)}{100}\bigg\rgroup\\\\\\:\implies\sf New\:Force=100+\bigg\lgroup 10 - 10 + ( - 1)\bigg\rgroup\\\\\\:\implies\sf New\:Force=100 - 1\\\\\\:\implies\sf New\:Force=99N$

⠀

$\therefore\:\underline{\textsf{Hence, new force of repulsion is \textbf{99N}}}.$