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

Correct answer: A. 4 π _0 Q × 10^ 22 ~V / m Key Idea : Search for the relations of electric potential and electric field at a particular point. At any point, electric potential due to charge Q is V= (1)/(4 π _0) · (Q)/(r) where r is the distance of observation point from the charge. At the same point, electric field is E= (1)/(4 π _0) · (Q)/(r^2) Combining Eqs. (i) and (ii), we have aligned E & =(4 π _0 V^2)/(Q)= 4 π _0 ×(Q × 10^ 11 )^2 Q & =4 π _0 Q ×10^ 22 ~V/ 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

Q

coulomb is

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

. The electric field at that point is

Options

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

Correct answer

A. $4 \pi \varepsilon_0 Q \times 10^{22} \mathrm{~V} / \mathrm{m}$

Step-by-step solution

Key Idea : Search for the relations of electric potential and electric field at a particular point. At any point, electric potential due to charge

Q

is V=

\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r}

where

r

is the distance of observation point from the charge. At the same point, electric field is E=

\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}

Combining Eqs. (i) and (ii), we have

\begin{aligned}E & =\frac{4 \pi \varepsilon_0 V^2}{Q}=\frac{4 \pi \varepsilon_0 \times\left(Q \times 10^{11}\right)^2}{Q}\\ & =4 \pi \varepsilon_0 \mathrm{Q} \times10^{22} \mathrm{~V}/ \mathrm{m} \end{aligned}

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