Question

A sphere of capacitance \(100\,\text{pF}\) is charged to a potential of \(100\,\text{V}\). Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is \(\alpha \times 10^{-7}\,\text{J}\). The value of \(\alpha\) is _____. (combined capacitance of spheres is \(200\,\text{pF}\)).

• A. \(5\)
• B. \( \frac{5}{2} \)
• C. \( \frac{7}{2} \)
• D. \( \frac{9}{2} \)

Answer 1

The Correct Option is B

Concept: Energy stored in a capacitor: \[ U=\frac{1}{2}CV^2 \] When identical spheres touch, charge redistributes equally and potential becomes half. Step 1: {Find initial energy.} \[ C=100\,\text{pF}=100\times10^{-12}\,\text{F} \] \[ V=100\,\text{V} \] \[ U_i=\frac12 CV^2 \] \[ =\frac12(100\times10^{-12})(100)^2 \] \[ =5\times10^{-7}\,\text{J} \] Step 2: {Find final potential.} Initial charge: \[ Q=CV \] \[ Q=100\times10^{-12}\times100 \] \[ =10^{-8}\,\text{C} \] When two identical spheres touch, charge divides equally: \[ Q'=\frac{Q}{2} \] Potential on each sphere: \[ V'=\frac{Q'}{C}=50\,\text{V} \] Step 3: {Find final energy.} Energy of each sphere: \[ U=\frac12 C V'^2 \] \[ =\frac12(100\times10^{-12})(50)^2 \] \[ =1.25\times10^{-7}\,\text{J} \] Total energy (two spheres): \[ U_f=2.5\times10^{-7}\,\text{J} \] Step 4: {Find change in energy.} \[ \Delta U=U_i-U_f \] \[ =5\times10^{-7}-2.5\times10^{-7} \] \[ =2.5\times10^{-7}\,\text{J} \] Thus \[ \alpha=\frac{5}{2} \]