The electric field strength due to ring of radius R at a distance x from its centre on the axis of ring carrying charge Q is given by - E - 14 - -160-0Qx-R-2 - x-2-3-2- At what distance from the centre will the electric field be maximum- x - R2- x - R-2- x - R- x - 2-R-

The correct option is B x = R√(2) #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10;E ...

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Standard XII

Physics

Infinite Line of Charge

The electric ...

Question

The electric field strength due to ring of radius R at a distance x from its centre on the axis of ring carrying charge Q is given by $E = \frac{1}{4 π ε_{0}} \frac{Q x}{(R^{2} + x^{2})^{3 / 2}}$ . At what distance from the centre will the electric field be maximum?

x = R

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x = \frac{R}{2}

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x = \frac{R}{\sqrt{2}}

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x = \frac{2}{\sqrt{R}}

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Solution

The correct option is B $x = \frac{R}{\sqrt{2}}$

E is maximum at a point where $\frac{d E}{d x} = 0$

$\frac{d E}{d x} = \frac{Q}{4 π ϵ_{0}} \frac{(x^{2} + R^{2})^{\frac{3}{2}} \cdot (1) - x \cdot \frac{3}{2} (x^{2} + R^{2})^{\frac{1}{2}} \cdot 2 x}{(x^{2} + R^{2})^{3}} = 0$

$⟹ (x^{2} + R^{2}) - 3 x^{2} = 0$

Thus, $R^{2} = 2 x^{2}$

$⟹ x = \pm \frac{R}{\sqrt{2}}$

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The electric field strength due to ring of radius R at a distance x from its centre on the axis of ring carrying charge Q is given by E=14πε0Qx(R2+x2)3/2. At what distance from the centre will the electric field be maximum?

The correct option is B x=R√2 E is maximum at a point where dEdx=0 dEdx=Q4πϵ0(x2+R2)32⋅(1)−x⋅32(x2+R2)12⋅2x(x2+R2)3=0 ⟹(x2+R2)−3x2=0 Thus, R2=2x2 ⟹x=±R√2

The electric field strength due to ring of radius R at a distance x from its centre on the axis of ring carrying charge Q is given by $E = \frac{1}{4 π ε_{0}} \frac{Q x}{(R^{2} + x^{2})^{3 / 2}}$ . At what distance from the centre will the electric field be maximum?