An uniform electric field E exists along positive x-axis. The work done in moving a charge $0.5 \mathrm{C}$ through a distance $2 \mathrm{~m}$ along a direction making an angle $60^{\circ}$ with $\mathrm{x}$ -axis is $10 \mathrm{~J}$. Then what is the magnitude of electric field (in…

Question

An uniform electric field E exists along positive x-axis. The work done in moving a charge $0.5 \mathrm{C}$ through a distance $2 \mathrm{~m}$ along a direction making an angle $60^{\circ}$ with $\mathrm{x}$ -axis is $10 \mathrm{~J}$. Then what is the magnitude of electric field (in $\mathrm{Vm}^{-1}$ )?

Aicharya · Accepted Answer

Force acting on the charged particle due to electric
field $=\mathrm{q} \overrightarrow{\mathrm{E}}$

work done in moving through distance $\mathrm{S}$,
$\mathrm{W}=\mathrm{q} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{S}}=(\mathrm{q} \mathrm{E}) \times \mathrm{S} \times \cos \theta$
$\therefore 10 \mathrm{~J}=(0.5 \mathrm{C}) \times \mathrm{E} \times 2 \cos 60^{\circ}$
$\mathrm{E}=10 \times 2=20 \mathrm{NC}^{-1}=20 \mathrm{Vm}^{-1}$

An uniform electric field E exists along positive x-axis. The work done in moving a charge $0.5 \mathrm{C}$…

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