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Electrostatics Physics Electrostatic Potential and Capacitance Physics Class 12th

In the diagram shown below, the block has charge $Q$ and is attached at one end of the light spring. The surface is smooth and horizontal and spring is in its natural length. If the electric field $E$ is switched on, the maximum elongation in the spring will be

Option 1 -

$\frac{Q E}{k}$

Option 2 -

$\frac{2 Q E}{k}$

Option 3 -

$\frac{Q E}{2 k}$

Option 4 -

$\frac{3 Q E}{2 k}$

2 Views | Posted a month ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

From work energy theorem

$W_{s p} + W_{E} = ? K$

$- \frac{1}{2} k x_{0}^{2} + Q E . x_{0} = 0$

$? x_{0} = \frac{2 Q E}{k}$

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In the diagram shown below, the block has charge $Q$ and is attached at one end of the light spring. The surface is smooth and horizontal and spring is in its natural length. If the electric field $E$ is switched on, the maximum elongation in the spring will be

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In the diagram shown below, the block has charge Q and is attached at one end of the light spring. The surface is smooth and horizontal and spring is in its natural length. If the electric field E is switched on, the maximum elongation in the spring will be

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In the diagram shown below, the block has charge $Q$ and is attached at one end of the light spring. The surface is smooth and horizontal and spring is in its natural length. If the electric field $E$ is switched on, the maximum elongation in the spring will be