Answer:
A. - 0.017N. It acts to the left.
B. - 0.043N. It acts to the left.
C. 0.060N. It acts to the right.
Explanation:
A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.
The net coulombs force on the charge is
F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)
Where K = Coloumbs constant =
Q(1) = charge on the leftmost side.
Q(2) = charge in the middle.
Q(3) = charge on the rightmost side.
F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)
F = 0.01753 - 0.03469
F = -0.017N
It has a negative sign, hence, it acts to the left.
B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.
The net coulombs force on the charge is
F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)
F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)
F = -0.017 - 0.02562
F = - 0.043N
It has a negative sign, hence, it acts to the left.
C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.
The net coulombs force on the charge is
F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)
F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)
F = +0.03469 + 0.02562
F = +0.060N
It has a positive sign, hence, it acts to the right.
