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Standard XII

Physics

Potential Energy - External Viewpoint

The plastic r...

Question

The plastic rod in the figure forms a half-circle and has charge $- Q$ uniformly distributed along its length $L$ . Find the electric field at the origin.

\frac{2 k Q}{π}

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\frac{2 k Q}{π L}

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\frac{2 k Q L}{π}

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2 k Q

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Solution

The correct option is B $\frac{2 k Q}{π}$

Half circle and charge $= - Q$ uniformly distributed along its length $= L$ , find the electric field
$E = \frac{μ_{0}}{4 π} \times \frac{Q}{L^{2}}$

$= \frac{{2 μ}_{0}}{π} \times \frac{Q}{L}$

$= \frac{2 k Q}{π L}$

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The plastic rod in the figure forms a half-circle and has charge −Q uniformly distributed along its length L. Find the electric field at the origin.

The correct option is B 2kQπHalf circle and charge =−Q uniformly distributed along its length =L, find the electric fieldE=μ04π×QL2 =2μ0π×QL =2kQπL

The plastic rod in the figure forms a half-circle and has charge $- Q$ uniformly distributed along its length $L$ . Find the electric field at the origin.

The correct option is B $\frac{2 k Q}{π}$

Half circle and charge $= - Q$ uniformly distributed along its length $= L$ , find the electric field
$E = \frac{μ_{0}}{4 π} \times \frac{Q}{L^{2}}$

$= \frac{{2 μ}_{0}}{π} \times \frac{Q}{L}$

$= \frac{2 k Q}{π L}$