An electric charge 10 - 6 μ C is placed at origin ( 0 , 0 ) m of X - Y co-ordinate system. Two points P and Q are situated at ( 3 , 3 ) m and ( 6 , 0 ) m respectively. The potential difference between the points P and Q will be :

The formula to calculate the potential difference between P and Q is given by ∆ V = k q 1 r 1 - 1 r 2 . . . 1 Now, r 1 = ( 3 ) 2 + ( 3 ) 2 m = 6 m And, r 2 = ( 6 ) 2 + 0 m = 6 m As r 1 = r 2 = 6 m and the value of the charge remains constant, the potential difference between the points is zero.

Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsPotential & potential energyHard2 minPYQ_2024

PhysicsHardsingle choice

An electric charge $10^{- 6} μ C$ is placed at origin $(0, 0)$ $m$ of $X - Y$ co-ordinate system. Two points $P$ and $Q$ are situated at $(\sqrt{3}, \sqrt{3}) m$ and $(\sqrt{6}, 0) m$ respectively. The potential difference between the points $P$ and $Q$ will be :

Options:

Answer:

Solution:

The formula to calculate the potential difference between $P$ and $Q$ is given by

$∆ V = k q (\frac{1}{r_{1}} - \frac{1}{r_{2}}) . . . (1)$

Now,

$\begin{array}{rcl} r_{1} & = & \sqrt{(\sqrt{3})^{2} + (\sqrt{3})^{2}} m \\ = & \sqrt{6} m \end{array}$

And,

$\begin{array}{rcl} r_{2} & = & \sqrt{(\sqrt{6})^{2} + 0} m \\ = & \sqrt{6} m \end{array}$

As $r_{1} = r_{2} = \sqrt{6} m$ and the value of the charge remains constant, the potential difference between the points is zero.

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Potential & potential energy

⏱ 2mℹ️ Source: PYQ_2024

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