An electric charge 10^ -3 μ C is placed at the origin (0,0) of (x-y) co-ordinate system. Two points A and B are situated at ( 2 , 2 ) and (2,0) respectively. The potential difference between the points A and B will be

Correct answer: C. zero aligned & r_1 = 2 i + 2 j & | r _1|=r_1= ( 2 )^2+( 2 )^2 =2 aligned aligned & r _2=2 i +0 j & or | r _2|=r_2=2 aligned Potential at point A is aligned & V_A=(1 q)/(4 π _0 r_1) & =(1)/(4 π _0) 10^ -3 × 10^ -6 2 aligned Potential at point B is aligned & V_B=(1)/(4 π _0) (q)/(r_2)=(1)/(4 π _0) 10^ -3 × 10^ -6 2 & V_A-V_B=0 . aligned

AIIMS 2010

Exam: AIIMS
Year: 2010
Subject: Physics
Chapter: Electrostatics

AIIMS 2010 · Physics · Electrostatics

An electric charge 10^ -3 μ C is placed at the origin (0,0) of...

An electric charge

10^{-3} \mu \mathrm{C}

is placed at the origin

(0,0)

(x-y)

co-ordinate system. Two points

A

and

B

are situated at

(\sqrt{2}, \sqrt{2})

and

(2,0)

respectively. The potential difference between the points

A

and

B

will be

An electric charge 10^ -3 μ C is placed at the origin (0,0) of (x-y) co-ordinate system. Two points A and B are

Options

A4.5 volt
B9 volt
Czero
D2 volt

Correct answer

C. zero

Step-by-step solution

\begin{aligned}& \overrightarrow{r_1}=\sqrt{2} \hat{i}+\sqrt{2} \hat{j}\\ & \left|\vec{r}_1\right|=r_1=\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=2 \end{aligned}

\begin{aligned}& \vec{r}_2=2 \hat{i}+0 \hat{j} \\ & \text{\text{ or } }\left|\vec{r}_2\right|=r_2=2 \end{aligned}

Potential at point

A

\begin{aligned}& V_A=\frac{1 q}{4 \pi \varepsilon_0 r_1}\\ & =\frac{1}{4 \pi \varepsilon_0} \frac{10^{-3} \times 10^{-6}}{2} \end{aligned}

Potential at point

is

\\

\begin{aligned}& \quad V_B=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r_2}=\frac{1}{4 \pi \varepsilon_0} \frac{10^{-3} \times 10^{-6}}{2}\\ & \therefore \quad V_A-V_B=0 . \end{aligned}

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