points) Problem S: Consider a (mostly) solid conducting sphere with radius, Ro.

Question

points) Problem S: Consider a (mostly) solid conducting sphere with radius, Ro. centered at the point O that has a net charge of O. The sphere has a circular cavity (empty space) centered at the point P. At P is a point charge of value +q. Matclh the region of space to the electric field in that region. (Hint: Use Gauss's Law) (4 points each) Ro o (5a) Inside the cavity? (5b) In the solid conductor? (Sc) Outside the conductor? Option a: q/(4ror) pointing radially towards O. Option b: ql(4nE2) pointing radially away from O. Option c: q/(46r2) pointing radially towards the P. Option d: g/(4ror2) pointing radially away from P Option e: 0. roblem 6: Find the power dissipated through each resistor in the following figure. (9 points) 100 10 VT -200 300

Explanation / Answer

a) E da = q/ ( epsilon)

E ( 4pi r^2) = q/ eqsioln

E = q / epsilon 4 pi r^2 ( directed away from P) ( option d)

b)  E da = q/ ( epsilon)

q = 0 inside any metallic surface

E = 0 ( option e)

c) E da = q/ epsilon

q= 0( no net charge )

E = 0 ( outside sphere) ( option e)

d) effective resistance= 100 + ( 200 x 300)/ 500 =220 ohms

I = V/R =10/ 220 =0.045 A

Power across 100 ohms= 0.045^2 (100) =0.2066 W

I across 200 ohms-=( 0.045) (300/ 500) = 0.027 A, Power across 200 ohms= 0.027^2 (200)=0.1458 P

I across 300 ohms= 0.018 A = power across 300 ohms= 0.018^2 (300)=0.0972 W

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