Answer:
[tex]q=2.997\times 10^{-4}[/tex]C
Sign-Negative
Explanation:
We are given that
Electric field =[tex]E=100NC^{-1}[/tex] (Radially downward)
Acceleration=[tex]0.19 ms^{-2}[/tex](Upward)
Mass of charge=3 g=[tex]3\times 10^{-3}[/tex]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
[tex]\sum F_y=ma[/tex]
[tex]\sum F_y=qE-mg[/tex]
Substitute the values then we get
[tex]qE-mg=ma[/tex]
Substitute the values then we get
[tex]q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)[/tex]
[tex]100q-29.4\times 10^{-3}=0.57\times 10^{-3}[/tex]
[tex]100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}[/tex]
[tex]q=\frac{29.97\times 10^{-3}}{100}[/tex]
[tex]q=2.997\times 10^{-4}[/tex]C
Sign of charge =Negative
Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.
