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An insulating solid sphere of the radius...

An insulating solid sphere of the radius R is charged in a non - uniform manner such that the volume charge density `rho=(A)/(r )`, where A is a positive constant and r is the distance from the centre. The potential difference between the centre and surface of the sphere is

A

`(AR)/(8epsilon_(0))`

B

`(AR)/(4epsilon_(0))`

C

`(AR)/(epsilon_(0))`

D

`(AR)/(2epsilon_(0))`

Text Solution

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The correct Answer is:
To find the potential difference between the center and the surface of an insulating solid sphere with a non-uniform charge density given by \(\rho = \frac{A}{r}\), we can follow these steps: ### Step 1: Understand the Charge Density The charge density \(\rho\) is given as \(\rho = \frac{A}{r}\), where \(A\) is a constant and \(r\) is the distance from the center of the sphere. This indicates that the charge density decreases as we move away from the center. ### Step 2: Calculate the Charge Element To find the electric field, we need to calculate the charge \(dq\) in a thin spherical shell of radius \(r\) and thickness \(dr\). The volume of this shell is: \[ dV = 4\pi r^2 dr \] Thus, the charge \(dq\) in this shell can be expressed as: \[ dq = \rho \cdot dV = \left(\frac{A}{r}\right) \cdot (4\pi r^2 dr) = 4\pi A r \, dr \] ### Step 3: Calculate the Total Charge Enclosed To find the electric field at a distance \(r\) from the center, we need to consider the total charge \(Q_{\text{enc}}\) enclosed within a sphere of radius \(r\): \[ Q_{\text{enc}} = \int_0^r dq = \int_0^r 4\pi A r' \, dr' = 4\pi A \left[\frac{(r')^2}{2}\right]_0^r = 2\pi A r^2 \] ### Step 4: Apply Gauss's Law Using Gauss's law, we know that the electric field \(E\) at a distance \(r\) from the center is given by: \[ \Phi = E \cdot A = \frac{Q_{\text{enc}}}{\epsilon_0} \] where \(A = 4\pi r^2\) is the surface area of the sphere. Thus, we have: \[ E \cdot 4\pi r^2 = \frac{2\pi A r^2}{\epsilon_0} \] From this, we can solve for \(E\): \[ E = \frac{2A}{4\epsilon_0} = \frac{A}{2\epsilon_0} \] ### Step 5: Calculate the Potential Difference The potential difference \(V\) between the center and the surface of the sphere can be calculated using the relationship: \[ \Delta V = -\int_{R}^{0} E \, dr \] Substituting the expression for \(E\): \[ \Delta V = -\int_{R}^{0} \frac{A}{2\epsilon_0} \, dr = -\left[\frac{A}{2\epsilon_0} r\right]_{R}^{0} = -\left(0 - \frac{AR}{2\epsilon_0}\right) = \frac{AR}{2\epsilon_0} \] ### Final Answer Thus, the potential difference between the center and the surface of the sphere is: \[ \Delta V = \frac{AR}{2\epsilon_0} \]
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