Question

what is the electric force acting between two charges of 0.0042 c and -0.0050 c that are 0.0030 m apart?

Answer 1

$\mathtt{ \huge{ \fbox{Solution :)}}}$

Given ,

The two charges are

• 0.0042 C or 42 × (10)^-4 C
• - 0.0050 C or 50 × (10)^-4 C

The distance between them is 0.0030 m or 30 × (10)^-4 m

We know that , the electrostatic force between two charges is given by

$\mathtt{\fbox{Electrostatic \: force = k \frac{ q_{1} q_{2} }{ {(r)}^{2} } }}$

Thus ,

$\sf \hookrightarrow Force = \frac{9 \times {(10)}^{9} \times 42 \times {(10)}^{ - 4} \times 50 \times {(10)}^{ - 4} }{ {(30 \times {(10)}^{ - 4}) }^{2} } \\ \\\sf \hookrightarrow Force = \frac{9 \times 2100 \times {(10)}^{(9 - 4 - 4 + 8)} }{900} \\ \\\sf \hookrightarrow Force = \frac{9 \times 21 \times {(10)}^{2} \times {(10)}^{9} }{9 \times {(10)}^{2} } \\ \\ \sf \hookrightarrow Force = 21 \times {(10)}^{9} \: \: newton$

Hence , the electrostatic force is 21 × (10)^9 newton

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