Question

What is the strength of an electric field that will balance the weight of a 1.0 g plastic sphere that has been charged to −3.0nc?

Answer 1

To balance the gravitational force on a 1.0 g plastic sphere with a –3.0 nC charge, an electric field strength of 3.27×106 N/C is needed. The calculation is based on setting the electric force equal to the gravitational force and solving for the electric field strength.

Step-by-step explanation:

The strength of an electric field that will balance the weight of a 1.0 g plastic sphere that has been charged to –3.0 nC can be computed using the relationship between the electric force and the weight of the sphere. The weight of the sphere is the force due to gravity acting on it, which is Fg = mg, where m is mass and g is the acceleration due to gravity (9.8 m/s2). To balance this force, the electric force Fe = qE, where q is the charge and E is the electric field strength, must be equal in magnitude to the gravitational force. Hence, solving for E we get:

E = Fg / q

E = (0.001 kg)(9.8 m/s2) / –3.0×10–9 C

E = –9.8×103 N/kg / –3.0×10–9 C

E = 3.27×106 N/C

The negative sign indicates that the direction of the electric field opposes the negative charge. However, in terms of magnitude, the electric field strength required to balance the weight of the plastic sphere is 3.27×106 N/C.

Answer 2

`3.27\cdot 10^6 V/m`

Step-by-step explanation:

In order to balance the weight of the sphere, the electric force must be equal in magnitude to the weight of the sphere:

`F_E = qE=mg`

where

`q=3.0nC=3.0\cdot 10^(-9) C` is the charge of the sphere (we can ignore the sign, since we are only interested in the magnitude of the force

E is the strength of the electric field

m = 1.0 g = 0.001 kg is the mass of the sphere

g = 9.81 m/s^2 is the gravitational acceleration

Solving the equation for E, we find the strength of the electric field:

`E=(mg)/(q)=((0.001 kg)(9.81 m/s^2))/(3.0\cdot 10^(-9) C)=3.27\cdot 10^6 V/m`