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Two identical thin ring, each of radius ...

Two identical thin ring, each of radius R meters, are coaxially placed a distance R metres apart. If `Q_1` coulomb, and `Q_2` coulomb, are repectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is

A

(a) zero

B

(b) `(q(Q_1-Q_2)(sqrt2-1))/((4sqrt2piepsilon_0R))`

C

(c) `(qsqrt2(Q_1+Q_2))/((4piepsilon_0R))`

D

(d) `(q(Q_1+Q_2)(sqrt2+1))/((4sqrt2piepsilon_0R))`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the work done in moving a charge \( q \) from the center of one ring to the center of the other ring, we can follow these steps: ### Step 1: Understand the Configuration We have two identical thin rings, each with radius \( R \), placed coaxially at a distance \( R \) apart. The charges \( Q_1 \) and \( Q_2 \) are uniformly distributed on the two rings. ### Step 2: Calculate the Electric Field at the Center of the Rings The electric field \( E \) at the center of a charged ring due to its own charge is zero because of symmetry. However, we need to calculate the electric field at the center of one ring due to the charge on the other ring. The distance from the center of one ring to the center of the other ring is \( R \). The electric field \( E \) at the center of ring 1 due to ring 2 can be calculated using the formula: \[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_2}{R^2} \] where \( \epsilon_0 \) is the permittivity of free space. ### Step 3: Calculate the Work Done The work done \( W \) in moving a charge \( q \) in an electric field \( E \) is given by: \[ W = q \cdot E \cdot d \] where \( d \) is the distance moved in the direction of the electric field. In this case, \( d = R \). Substituting the expression for \( E \): \[ W = q \cdot \left(\frac{1}{4\pi \epsilon_0} \cdot \frac{Q_2}{R^2}\right) \cdot R \] This simplifies to: \[ W = \frac{q \cdot Q_2}{4\pi \epsilon_0 R} \] ### Final Result Thus, the work done in moving the charge \( q \) from the center of one ring to the center of the other ring is: \[ W = \frac{q \cdot Q_2}{4\pi \epsilon_0 R} \]

To solve the problem of finding the work done in moving a charge \( q \) from the center of one ring to the center of the other ring, we can follow these steps: ### Step 1: Understand the Configuration We have two identical thin rings, each with radius \( R \), placed coaxially at a distance \( R \) apart. The charges \( Q_1 \) and \( Q_2 \) are uniformly distributed on the two rings. ### Step 2: Calculate the Electric Field at the Center of the Rings The electric field \( E \) at the center of a charged ring due to its own charge is zero because of symmetry. However, we need to calculate the electric field at the center of one ring due to the charge on the other ring. ...
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