Question

The spherical surface r = 1 m, 2 m and 3 m carry surface charge densities of 20, -9 and 17. 2 nC/m2 respectively Calculate the electric flux leaving through the surface r 5 m а. Find electric flux density at P (1, -1, 2) b. Four charges are located at the vertices of a rectangular plane shown in Fig. Find the magnitude and direction of resultant force on Qı. The width of the plane is 5 cm and the height of the plane is 2 cm.

Answer 1

weng Gaur's law , the tobal charge in the closed surface is given by Q- (hr. ds Sg (4782) Total charge in spherical pero face of radius, 8.1m is quien by: 8. S (4887) Q, = zo x 169 (UNCR) - 251.32x1099 P-261mic Similarly for R=2 Q = Is (4882) Q2 = 9x109) (4a (2/2) O2 = - 452.38 mc Similarly for R=3 Uz la CUA OR?) - 24109 (470)2) = 226.2 nc How total charge in spher 8= 5m is given by cal surface of radius, 0 = 8, + Q2 + Q₂ - 251.3 -45213 +22612 Q = 25.11 nc

By Gauss's law , total charge in closed serfare a quien by = foods a = D(4882) ao Now charge density in at 8=5m is given by : spherical perfare YA82 95.11109 a un (57 - 0.0799 x 159 ar chuir = 0.08 Ar nelam? At Pll,-1, 2) D-0.08 C1 - / + 2k) melain = 0.08 (1-2 +22) 2.45 =(o. 032 - 0. maj + 0.064E) nClann

Now y vertices the Four of a Charges are rectangular pland at plane - aga 5cm The diagonal distame blow blue charges are by 2² +52 = S:38cm. Now force experienced on 9. due to other charges are quien by F = 9, Q2 + Q & Q Qu Lato 20.06) Yato(0.02)2 Unto (0.053012 Direction will be charges whether depend they are on the positive negative. Š e anagnitude of charge or positive cholen

Answer 2

Given data The first radius of the spherical surface is, 11 = 1m The second radius of the spherical surface is, 11 = 2 m The third radius of the spherical surface is, z = 3 m The value of the first charge density is, Ps1 = 20 nC/m² The value of the second charge density is, Ps2 =-9 nC/m? The value of the third charge density is, Psa = 2 nC/m? The radius of the surface from which the electric flux is leaving is, r = 5 m (a) Write the expression for the gauss law. Q = P.-ds = P: (41r?) Write the expression for the total charge in the spherical surface of radius r =1m. Q = P. (477?) Substitute values in the above equation. Q 20 nC/m (4 (1m)^) = 251.3 nc

Write the expression for the total charge in the spherical surface of radius r = 2 m. Q2 = P.2 (4trz) Substitute values in the above equation. Q2 = -9nC/mº (47(2 m)) = -452.38 nc Write the expression for the total charge in the spherical surface of radius r = 3 m. Q = Ps (4trz) Substitute values in the above equation. Q: = 2 nC/m² (47(3 m)) = 226.2 nC Write the expression for the total charge of the spherical surface of radius r = 5m. Q=Q+Q+Q:

Substitute the values in the above equation. Q = 251.3 nC-452.3+226.2 nC, = 25.11nC Write the expression for the total charge in the closed surface. Q=D ( 4rtra, D- (audio) Substitute values in the above equation. D- 25.11nC a (41(5 m)) = 0.08a, nC/m2 Hence, the expression for the flux leaving from the surface of radius r = 5 m is 0.08a, nC/m.

(b) Write the expression for the surface charge density at points (1,-1,2). nor îf û+ 2û 11? +12 +22" =(0.032i – 0.032j+0.064k) nC/m2 Hence, the value of the electric flux density is (0.032i-0.032j+0.064k) nC/mº.