Question

A force of 4•5 works on a charge of 5×10^-4c place at any pont. find electric field at that point

Answer 1

$\color{darkblue}\underline{\underline{\sf Given-}}$

• Force (F) = 4.5 N
• Charge (Q) = ${\sf 5×10^{-4}C}$

$\color{darkblue}\underline{\underline{\sf To \:Find-}}$

• Electric Field (E)

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$\color{green}\underline{\underline{\sf Formula\: Used-}}$

$\color{violet}\blacksquare\underline{\boxed{\sf E=\dfrac{F}{Q}}}$

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$\implies{\sf E=\dfrac{4.5}{5×10^{-4}}}$

$\implies{\sf E=0.9×10^{4}N/C}$

$\color{red}\implies{\sf E=9×10^3\:N/C}$

$\color{darkblue}\underline{\underline{\sf Answer-}}$

Electric Field will be $\color{red}{\sf 9×10^3\:N/C}$.

Answer 2

Given ,

• Electrostatic force (f) = 4.5 Newton
• Charge (q) = 5 × (10)^-4 C

We know that ,

The electric field is defined as the electrostatic force per unit charge

$\large \mathtt{ \fbox{Electric \: field = \frac{f}{q} }}$

Thus ,

$\sf \mapsto Electric \: field = \frac{4.5}{5 \times {(10)}^{ - 4} } \\ \\ \sf \mapsto Electric \: field = \frac{45}{5 \times 10 \times {(10)}^{ - 4} } \\ \\ \sf \mapsto Electric \: field = \frac{9}{ {(10)}^{ - 3} } \\ \\ \sf \mapsto Electric \: field = 9 \times {(10)}^{3} \: \: n/c$

Hence , the electric field is 9 × (10)^3 n/c

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