Question

A charge Q is distributed uniformly on a fixed ring of radius a. A point charge q is placed on axis at a distance 2a from the centre of ring. The force experienced by point charge at given position is?????

Answer 1

Given,

A charge Q is distributed uniformly on a fixed ring of radius a. A point charge q is placed on axis at a distance 2a from the centre of ring.

To find,

The force experienced by point charge at given position.

we know, electric field due to ring at a point lies along the axis of ring is given by,

E = KQx/(x² + R²)³/²

where x is the seperation between centre and observation point, R is the radius of ring and Q is the charge distributed uniformly on the ring.

here, x = 2a, R = a

then, E = KQ(2a)/(4a² + a²)³/²

= 2KQa/5√5a³

= 2KQ/5√5a²

now force experienced by point charge at given position, F = qE

= 2KQq/5√5a²

Therefore force experienced by point charge at given position is 2KQq/5√5a²

Answer 2

Answer:

• Force (F) experienced is 2 K Q q / 5 √{5} a²

Given:

1. Radius of ring (R) = a
2. Distance from ring (x) = 2 a

Explanation:

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Firstly we need to find the Electric field on charge "q"

From the formula, we know,

$\bigstar\;\underline{\boxed{\sf E=\dfrac{K\;Q\;x}{\bigg(x^2+R^2\bigg)^{3/2}}}}$

Here,

• K Denotes constant.
• Q Denotes Charge on ring.
• x Denotes distance of separation.
• R Denotes radius of ring.

Substituting the values,

$\displaystyle\longrightarrow\sf E=\dfrac{K\;Q\times \Big(2\;a\Big)}{\bigg(\Big\langle2\;a\Big\rangle^2+\Big\langle a\Big\rangle^2\bigg)^{3/2}}\\\\\\\longrightarrow\sf E=\dfrac{2\;K\;Qa}{\bigg( 4\;a^{2}+a^{2}\bigg)^{3/2}}\\\\\\\longrightarrow\sf E=\dfrac{2\;K\;Qa}{\bigg( 5\;a^{2}\bigg)^{3/2}}$

• $\sf 5\;a^{2}=\bigg\langle \sqrt{5}\;a \bigg\rangle^{2}$

$\displaystyle\longrightarrow\sf E=\dfrac{2\;K\;Qa}{\bigg( \Big\langle \sqrt{5} \; a \Big\rangle^{2} \bigg)^{3/2}}\\\\\\\longrightarrow\sf E=\dfrac{2\;K\;Qa}{\bigg(\sqrt{5}\;a\bigg)^{2\;\times \;3/2}}\\\\\\\longrightarrow\sf E=\dfrac{2\;K\;Qa}{\bigg(\sqrt{5}\;a\bigg)^{3}}$

• Cubing the denominator.

$\displaystyle\longrightarrow\sf E=\dfrac{2\;K\;Qa}{5\sqrt{5}\;a^{3}}\\\\\\\longrightarrow\sf E=\dfrac{2\;K\;Q}{5\sqrt{5}\;a^{2}}\\\\\\\longrightarrow \underline{\boxed{\sf E=\dfrac{2\;K\;Q}{5\sqrt{5}\;a^{2}}}}$

∴ We got the Electric field.

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From the formula, we know,

$\bigstar\;\underline{\boxed{\sf F=q\;E }}$

Here,

• F Denotes Force.
• q Denotes charge.
• E Denotes Electric field.

Substituting the values,

$\displaystyle\longrightarrow\sf F=q\times \dfrac{2\;K\;Q}{5\sqrt{5}\;a^{2}}\\\\\\\longrightarrow\sf F=\dfrac{2\;K\;Q\;q}{5\sqrt{5}\;a^{2}}\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf F=\dfrac{2\;K\;Q\;q}{5\sqrt{5}\;a^{2}}}}}}$

∴ We got the Force value.

For more clarity, Refer the attachment.

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