Question

Near the surface of Earth, an electric field points radially downward and has a magnitude of approximately 100 N/C. 1) What charge (magnitude and sign) would have to be placed on a penny that has a mass of 2.50 g to cause it to rise into the air with an upward acceleration of 0.190 m/s2? (Express your answer to three significant figures.)

Answer 1

`-2.50*10^(-4) \text{ C}` or -0.000250 C

Step-by-step explanation:

The field points downward, which is the direction of a positive charge. In order for the penny to rise, it has to have a negative charge since its direction is opposite that of the field.

To calculate the magnitude of the charge:

The penny is to accelerate upward so it must overcome gravity. Hence, the net force to cause it to accelerate it upward is the difference between the electrostatic force and its weight.

`F_N = F_E - W`

`F_E = F_N + W`

`F_E = ma + mg`

`F_E = m(a + g)`

`F_E = 0.0025(0.19+9.8)`

`F_E = 0.0025*9.99`

Now, the electric field intensity is given by

`F_E = E * q`

where q = charge

`q=(F_E)/(E)`

`q=(0.0025*9.99)/(100)=0.00024975= 0.000250` to 3 significant figures.

In standard form, q =
`-2.50*10^(-4) \text{ C}` The negative sign indicates it has a negative charge, as explained initially.