Ahollow, conducting sphere with an outer radius of 0.240 m and an inner radius of 0.200 m has a uniform surface charge density of +6.67 × 10^−6 c/m^2. a charge of -0.600 μc is now introduced into the cavity inside the sphere. calculate the strength of the electric field at a distance 0.11 m from the center of the sphere.

446.28*10^3 N/C ( directed to center of the sphere)

Explanation: In order to explain this problem we have to use the electric field expression, which is given by:

E=k*q/r^2 where k is a constant equal 9*10^9 N/C^2*m^2, r is the distance from the charge to calculate the electric field.

Then we have,

E=(9*10^9*0.6*10^-6)/(0.11)^2= 446.28*10^3 N/C

Due to the negative point charge, the electric field is directed to the center of the sphere.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.