Question

 

5). How many electrons have to be added to 1 milligram metal sphere such that another +1C charge located 12 mm above the sphere will be able to hold that in the air?[Assume g=10m/s^2].

6). Two point charges, q₁= +13 nC and q₂= -7 nC, are located on the x-axis at x= 0 and x=9 m. What is the distance of ZERO electric field from the positive charge?

7). In a right angle, AB = 2 m, and angle ACB is 63.43 Degree. A point charge of 2* 23 nC is placed at point A and another point charge -3* 23 nC is placed at point C. Calculate the POTENTIAL at B

8). Two charges, one is at A with - 30.44 nC and other is at B with +9* 30.44 nC are seperated by 1 m. Find the distance AC in cm for which electric POTENTIAL at point C is zero. Point C is located on line AB.

Answer 1

5) let m = 1 milligram = 1*10^-6 kg
q1 = 1 C
d = 12 mm = 0.012 m
let q2 is the charge added to the metal sphere

in the equilibrium, Fe = Fe

k*q1*q2/d^2 = m*g

q2 = m*g*d^2/(k*q1)

= 1*10^-6*10*0.012^2/(9*10^9*1)

= 1.6*10^-19 C

number of electrons added, N = q2/e

= 1.6*10^-19/(1.6*10^-19)

= 1.0000 <<<<<----------Answer

6) let d is the distance from positive charge where electric field is zero.

so, at a distance d from the positive charge, |E1| = |E2|

k*|q1|/d^2 = k*|q2|/(d-x)^2

|q1|/d^2 = |q2|/(d-x)^2

13/d^2 = 7/(d-9)^2

==> d = 33.8091 m <<<<<----------Answer

7) from the given data, BC = AB/sin(63.43)

= 2/sin(63.43)

= 2.236 m

potential at B = k*qA/AB + k*qC/BC

= 9*10^9*2*23*10^-9/2 + 9*10^9*(-3*23)*10^-9/2.236

= -70.7281 V <<<<<----------Answer

8) let x is the distance from point A where electric potential is zero.

so, at a distance x from point A,

k*|qA|/x = k*|qB|/(1 - x)

|qA|/x = |qB|/(1 - x)

30.44*10^-9/x = 9*30.44*10^-9/(1 - x)

==> x = 0.1000 m <<<<<<<<<<-------------------Answer