Question

If the given figure shows equipotential surfaces then the magnitude of electric field is

• A. $100{\text{NC}}^{-1}$
• B. $50{\text{NC}}^{-1}$
• C. $400{\text{NC}}^{-1}$
• D. $200{\text{NC}}^{-1}$

Answer 1

The correct option is C $400{\text{NC}}^{-1}$



From the figure we get,

$\sin 30=\frac{BC}{AC}=\frac{d}{10}$

$\therefore d=10\sin {30}^{\circ }=10\times \frac{1}{2}=5\text{cm}$

$E=\frac{\Delta V}{d}=\frac{40-20}{5\times {10}^{-2}}=\frac{20}{5\times {10}^{-2}}=400{\text{NC}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (c) is the correct answer.