Saved Bookmarks
| 1. |
Two charged conducting spheres of radii a and b are connected to each other by a wire. What is theratio of electric fields at the surfaces of the two spheres ? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions. |
|
Answer» SOLUTION :When two charged conducting spheres of RADII a and b are connected to each other by a wire, theyshare their charges and acquire same potential i.e., `V_1 = V_2`. If charges on two spheres be `Q_1 and Q_2`respectively, then it means that `Q_1/(4pi epsi_0a) = Q_2/(4pi epsi_0b) or (Q_1)/Q_2 = a/b` Now electric fields at the surfaces of two spheres will have a ratio `E_1/E_2 = (Q_1/(4pi epsi_0a^2))/(Q_2/(4pi epsi_0b^2)) = Q_1/(Q_2).b^2/a^2 = a/b.b^2/a^2 =b/a` As electric FIELD very near a conducting surface is given by `E = sigma/epsi_0`, hence surface charge densities will have the ratio `sigma_1/sigma_2 = E_1/E_2 = b/a` Thus, surface density of charge at a point of a conductor is inversely proportional to the radius (or THICKNESS). Hence, charge density of a conductor is much higher at the sharp and POINTED ends andless on its flatter portion. |
|