Question

Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (magnitude and sign) would have to be placed on a penny that has a mass of 3g to cause it to rise into the air with an upqard acceleration of 0.19 m/s^2

Answer 1

`q=2.997* 10^(-4)`C

Sign-Negative

Step-by-step explanation:

We are given that

Electric field =
`E=100NC^(-1)` (Radially downward)

Acceleration=
`0.19 ms^(-2)`(Upward)

Mass of charge=3 g=
`3* 10^(-3)`kg

1kg=1000g

We have to find the magnitude and sign of charge would have to be placed on a penny .

By newton's second law

`\sum F_y=ma`

`\sum F_y=qE-mg`

Substitute the values then we get

`qE-mg=ma`

Substitute the values then we get

`q(100)-3* 10^(-3)(9.8)=3* 10^(-3)(0.19)`

`100q-29.4* 10^(-3)=0.57* 10^(-3)`

`100q=0.57* 10^(-3)+29.4* 10^(-3)=29.97* 10^(-3)`

`q=(29.97* 10^(-3))/(100)`

`q=2.997* 10^(-4)`C

Sign of charge =Negative

Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.