Full Question
Consider two point charges located on the x axis: one charge, Q1 = -12.0 nC , is located at x1 = -1.705m ; the second charge, Q2 = 36.5 nC, is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Your answer may be positive or negative, depending on the direction of the force.
Answer:
6.79E6 N
Explanation:
Given
Q1 is negative and to the left of Q3 the force will be to the left
Q2 is positive and to the right of Q3 the the force will also be to the left
Net Force is calculated as:
Using Coulomb's law
Coulomb's law: F = kqQ / r²
the constant k = 8.99 x 10^9 N m2 / C2
F = -kQ1*Q3/(r1)² -kQ2*Q3/(r2)²)
F = -kQ3(Q1/(r1)² + Q2/(r2)²)
Where
Q1 = -12nC = -12 * 10^-9C
Q2 = 36.5nC = 36.5 * 10^-9C
Q3 = 49.5nC = 49.5 * 10^-9C
x1 = -1.705m
x2 = x = 0
x3 = -1.170m
r1 = x3 - x1
r1 = -1.170 - -1.705
r1 = -1.170 + 1.705
r1 = 0.535
r1² = 0.286225
r2 = x3
r2 = -1.170
r2² = -1.170²
r2² = 1.3689
So,
F = (-8.99 * 10^9)(49.5 *10^-9) [-12 * 10^-9/0.286225 + 36.5 * 10^-9/1.3689]
F = -445.005 (−4.192505895711E−8 + 2.6663744612462E−8)
F = -445.005 * −1.5261314344648E−8
F = -(8.99 * 10^9) * (49.5 * 10^-9) * [ (-12 * 10^-9) /(-1.770 - -1.705)² + (36.5 * 10^-9)/(-1.170)²]
F = -445.005( −0.000002813572941777538)
F = 0.00000679136118994008324
F = 6.79E6 N
