The electric potential at a point in free space due to a charge Q coulomb is Q × 10^ 11 volts. The electric field at that point is

Correct answer: B. 4 π _0 Q × 10^ 22 volt/m aligned V & =(1)/(4 π _0) · (Q)/(R)=Q ×10^ 11 volt E & =(1)/(4 π _0) · (Q)/(R^2)=(V)/(R)=Q ×10^ 11 ×4 π _0 ×10^ 11 [ from ...(i) ] & =4 π _0 Q ×10^ 22 volt / m aligned

NEET 2008

Exam: NEET
Year: 2008
Subject: Physics
Chapter: Electrostatics

NEET 2008 · Physics · Electrostatics

The electric potential at a point in free space due to a charge Q...

The electric potential at a point in free space due to a charge

\mathrm{Q}

coulomb is

\mathrm{Q} \times 10^{11}

volts. The electric field at that point is

Options

A $12 \pi \varepsilon_0 Q \times 10^{22}$ volt/m
B $4 \pi \varepsilon_0 Q \times 10^{22}$ volt/m
C $12 \pi \varepsilon_0 \mathrm{Q} \times 10^{20} \mathrm{volt} / \mathrm{m}$
D $4 \pi \varepsilon_0 \mathrm{Q} \times 10^{20}$ volt/m

Correct answer

B. $4 \pi \varepsilon_0 Q \times 10^{22}$ volt/m

Step-by-step solution

\begin{aligned}V & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R}=Q \times10^{11} \text{ volt }\\ E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R^2}=\frac{V}{R}=Q \times10^{11} \times4 \pi \varepsilon_0 \times10^{11}[\text{ from ...(i) }]\\ & =4 \pi \varepsilon_0 Q \times10^{22} \text{ volt }/ \mathrm{m} \end{aligned}

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