Question

A charge of -0.30 C is placed on a circular conducting plate (very thin) with radius R= 10 cm. a) Draw a diagram of the situation described above, and calculate the surface charge density o. b) What is the magnitude of the electric field at a point 3.0 cm above the center of the plate? c) What is the magnitude of the electric field at that point if a 5 cm hole is placed in the center of the disk. Hint: This is a tricky one. Use the principle of superposition: this is equivalent to having a 5cm radius disk of equal POSITIVE charge in the middle of your disk of negative charge, so that the charges cancel.

Answer 1

The surface charge density is: = Q πα -0.30 A(0.10) = -9.5 C/m2 2 a b) The electric field at the given distance is:

N 1 - E= 260 VpP+R? 0.03 1 - -9.5 2(8.85x10-12) = -3.8×10IN/ 0.03 +0.107 c)The electric field at the given distance is: N .X E= 1 260 1- V2 + R2 280 vr + R2 +11 - 부 0.03 0.03 +0.10 9.5 2(8.85×10-12 = -1.2×10 N/C 0.03 0.03 + 0.05