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

Answer:

$q=2.997\times 10^{-4}$C

Sign-Negative

Explanation:

We are given that

Electric field =$E=100NC^{-1}$ (Radially downward)

Acceleration=$0.19 ms^{-2}$(Upward)

Mass of charge=3 g=$3\times 10^{-3}$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

$q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)$

$100q-29.4\times 10^{-3}=0.57\times 10^{-3}$

$100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}$

$q=\frac{29.97\times 10^{-3}}{100}$

$q=2.997\times 10^{-4}$C

Sign of charge =Negative

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