Question

Two identical small plastic balls are rubbed against a piece of wool such that both of them received approximately 2×10^12 and 2.5×10^12 electrons respectively. now both the charged balls are kept at distance of 25cm in air. calculate the nature and magnitude of force between both the balls.

​

Answer 1

Answer :-

$\to \boxed{\sf{{F} = 7.2 \times {10}^{35} } \: N \: } \:$

Nature → Repulsive force .

To Find :-

→ Nature and magnitude of force between them .

Explanation :-

According to the question

Given charges are

$\star \: \sf{q_{1} = 2 \times {10}^{12} } \: c \\ \star \: \sf{q_{2} = 2.5 \times {10}^{12} } \\ \star \: \sf{ distance \: (r) = 25cm = 0.25 \: m \: }$

We know that

$\implies \boxed{\sf{ {F} = \frac{1}{4\pi\epsilon_{o} } \: \frac{q_{1}q_{2}}{ {r}^{2} } }} \\ \because \sf{ \frac{1}{4\pi\epsilon_{o} } = 9 \times {10}^{9} }$

substitute the given values

$\sf{ \to{F} = 9 \times {10}^{9} \times \frac{2 \times 2.5 \times {10}^{24} }{0.25 \times 0.25} } \\ \\ \to \sf{{F} = \frac{9 \times 2 \times 25}{25 \times 25} \times {10}^{27 + 9} } \\ \\ \to{F} = \frac{180}{25} \times {10}^{35} \\ \\ \to \boxed{\sf{{F} = 7.2 \times {10}^{35} } \: N \: }$

this is the required force.

Nature is Repulsive force Because like charges repel and unlike charges attract to each other .here both charges are positive hence force is repulsive.

Answer 2

$\tt{\orange {\huge {Hello!!! }}} s.ld$

QUESTION :

Two identical small plastic balls are rubbed against a piece of wool such that both of them received approximately 2×10^12 and 2.5×10^12 electrons respectively.

Now both the charged balls are kept at distance of 25cm in air.

Calculate the nature and magnitude of force between both the balls.

SOLUTION :

The following information has been provided...

The magnitude of the two charges are 2×10^12 and 2.5×10^12 respectively.

The distance between these isolated charges is 0.25 m.

$\begin{lgathered}\leadsto \purple{\sf{ {F} = \frac{1}{4\pi\epsilon_{o} } \: \frac{q_{1}q_{2}}{ {r}^{2} } }} \\ \because \sf{ \frac{1}{4\pi\epsilon_{o} } = 9 \times {10}^{9} } \\\end{lgathered}$

Now we have to substitute the above values in the equation.

Substituting ,

$\begin{lgathered}\sf \green{ \to{F} = 9 \times {10}^{9} \times \frac{2 \times 2.5 \times {10}^{24} }{0.25 \times 0.25} } \\ \\ \to \sf{{F} = \frac{9 \times 2 \times 25}{25 \times 25} \times {10}^{27 + 9} } \\ \\ \to{F} = \frac{180}{25} \times {10}^{35} \\ \\ \to \blue \mapsto{\sf{{F} = 7.2 \times {10}^{35} } \: N \: }...(A)\end{lgathered}$