Question

What is the magnitude of an electric field that balances the weight of a plastic sphere of mass 8.5 g that has been charged to -3.0 nC?

A) 8.5×10^6 N/C
B) 3.1×10^6 N/C
C) 6×10^6 N/C
D) 2.8×10^7 N/C

Answer 1

The magnitude of the electric field that balances the weight of a charged plastic sphere of mass 8.5 g and charge -3.0 nC is approximately 2.8×10^7 N/C.

Step-by-step explanation:

The magnitude of the electric field that balances the weight of a charged plastic sphere can be found using the formula F = qE, where F is the weight of the sphere, q is the charge, and E is the electric field strength. In this case, the weight of the sphere is equal to its mass times the acceleration due to gravity (F = mg). The charge of the sphere can be converted from nC to C by dividing by 10^9. Plugging in the values, we have:

F = (0.0085 kg)(9.8 m/s^2) = 0.0833 N

E = F/q = (0.0833 N)/(-3x10^-9 C) ≈ -2.8x10^7 N/C

Therefore, the magnitude of the electric field that balances the weight of the sphere is approximately 2.8×10^7 N/C. The correct answer is (D).

Answer 2

Approximately
`2.8 * 10^(7)\; {\rm N \cdot C^(-1)}`, assuming that
`g = 9.81 \; {\rm N\cdot kg^(-1)}`.

Step-by-step explanation:

Assuming that gravitational attraction and the electrostatic force from the electric field are the only forces on this sphere. The magnitude of this electric field can be found in the following steps:

• Find the gravitational force the plastic sphere.
• Deduce that the electrostatic force on the sphere is equal in magnitude to the gravitational force on the sphere.
• Divide the magnitude of the electrostatic force on the sphere by the magnitude of the charge on the sphere to find the magnitude of the electric field.

It is given that the mass of this sphere is
`m = 8.5\; {\rm g}`. Apply unit conversion and ensure that mass is measured in standard units (kilograms):

`\begin{aligned} m = 8.5\; {\rm g} = 8.5 * 10^(-3)\; {\rm kg} \end{aligned}`.

The magnitude of the weight of the sphere (gravitational force from the planet) would be:

`\begin{aligned} (\text{weight}) &= m\, g\end{aligned}`.

Assume that that the only two forces on this sphere are gravitational attraction and electrostatic force. Since the forces on the sphere to be balanced, the magnitude of the electrostatic force should be equal to that of the gravitational attraction on the sphere,
`m\, g`.

In this question, it is given that the electrostatic charge on this sphere is
`(-3.0)\; {\rm nC}` (magnitude of the charge is
`3.0\; {\rm nC}`.) Apply unit conversion and ensure that the magnitude of the charge is measured in the standard unit, Coulomb (instead of nanocoulombs):

`\begin{aligned}q &= 3.0\; {\rm nC} = 3.0 * 10^(-9)\; {\rm C}\end{aligned}`.

Divide the magnitude of the electrostatic force by the charge on the sphere to find the strength of the electric field:

`\begin{aligned}E &= \frac{(\text{electrostatic force})}{(\text{electrostatic charge})} \\ &= (m\, g)/(q) \\ &= \frac{(8.5 * 10^(-3)\; {\rm N})\, (9.81\; {\rm N\cdot kg^(-1)})}{(3.0* 10^(-9)\; {\rm C})}\\ &\approx 2.8 * 10^(7)\; {\rm N\cdot C^(-1)}\end{aligned}`.

In other words, under the assumptions, the magnitude of the electric field would be approximately
`2.8 * 10^(7)\; {\rm C}`.