Location A is 3.40 m to the right of a point charge q. Location B lies on the same line and is 4.50 m to the right of the charge. The potential difference between the two locations is VB - VA = 45.0 V. What is the magnitude and sign of the charge?
ΔV=VB-VA =kq/r(B)-kq/r(A)=
=kq{r(a)-r(B)}/r(A)•r(B).
q = r(A)•r(B)•ΔV/k{r(a)-r(B)}=
=(3.4-4.5)•45/9•10⁹•3.4•4.5 =
= - 9.5•10⁻¹⁰ C
To find the magnitude and sign of the charge, we can use the equation for potential difference (V):
V = k * (q / r),
where V is the potential difference, k is Coulomb's constant (8.99 x 10^9 N⋅m^2/C^2), q is the charge, and r is the distance.
In this case, we have the potential difference (VB - VA) = 45.0 V, the distance from the charge to location A (rA) = 3.40 m, and the distance from the charge to location B (rB) = 4.50 m.
First, we need to calculate the potential at each location:
VA = k * (q / rA),
VB = k * (q / rB).
Next, we substitute these values into the equation for potential difference:
VB - VA = k * (q / rB) - k * (q / rA).
Rearranging the equation, we get:
(q / rB) - (q / rA) = (VB - VA) / k.
Now, we can substitute the given values:
(1 / 4.50) - (1 / 3.40) = (45.0 V) / (8.99 x 10^9 N⋅m^2/C^2).
Simplifying the equation gives:
(1 / 4.50) - (1 / 3.40) = 5.005562845 x 10^-8 C.
Now, we can solve for q:
q = [(1 / 4.50) - (1 / 3.40)] * 5.005562845 x 10^-8 C.
Evaluating this expression, we find the magnitude and sign of the charge.