Question

The force of attraction between two identical charges is 60 N. If the amount of charge is increased to twice of its initial, what will be the force now?​

Answer 1

Answer:

Let the distance between the charges initially be r.

Thus initial electrostatic force F

o

=

r

2

kQ

1

Q

2

New distance between the charges r

′

=3r

Charge of each particle is tripled Q

1

′

=3Q

1

and Q

2

′

=3Q

2

∴ New electrostatic force F

′

=

(3r)

2

k(3Q

1

)(3Q

2

)

=F

o

Answer 2

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Given, Initial Force of att. = 60 N.

Let the two charges be q₁ and q₂ respectively.

So, Now from the coulombs law we know that,

$\longrightarrow\sf F=\dfrac{1}{4 \pi \epsilon_{0}}\;.\; \dfrac{q_{1}q_{2}}{r^{2}}$

So,

$\longrightarrow\sf F \propto q_{1}\;.\; q_{2}$

As now in the second case both the charge is doubled to the initial value, lets find the force of attraction in the second case,

$\begin{lgathered}\longrightarrow\sf \dfrac{F_{1}}{F_{2}} = \dfrac{q_{1}\;.\; q_{2}}{q_{3}\;.\; q_{4}}\\\\\\\\\longrightarrow\sf \dfrac{60}{F_{2}}=\dfrac{q_{1}\;.\; q_{2}}{2\;q_{1}\;.\; 2\; q_{2}}\\\\\\\\\longrightarrow\sf \dfrac{60}{F_{2}}=\dfrac{q_{1}\;.\; q_{2}}{4\;(q_{1}\;.\;q_{2})}\\\\\\\\\longrightarrow\sf \dfrac{60}{F_{2}}=\dfrac{1}{4}\\\\\\\\\longrightarrow\sf F_{2} = 60\times 4\\\\\\\\\longrightarrow\sf F_{2}=240 N\end{lgathered}$

$\longrightarrow\sf\frac{60}F_{2}= \frac{1}{4}$

$\sf{F_{2} = 60 \times 4}$

$\longrightarrow \sf{F_{2} = 240N}$

∴ The Force between the charge's is 240 N.