r1 =r3 = 1cm = 10^-2 m
The Pythagorean Theorem gives the distance from the midpoint of the base to the charge at the apex of the triangle as
r2 = sqrt(4² - 1²) = sqrt(15) = 3.87 cm = 3.87•10^-2 m.
The potential at the midpoint of the base is
φ = k•q1/r1 + k•q2/r2 + k•q3/r3 = k{(q1+q3)/r1+ q2/r2} =
= 9•10^9• {(- 16•10^-9 - 16•10^-9)/10^-2 +16•10^-9/3.87•10^-2} =
= 9•10^9•(-32•10^-7 +4.13•10^-7)= - 2.5•10^5 V = - 25 kV
The three charges shown in Figure P16.17 are at the vertices of an isosceles triangle. If q = 16.0 10-9 C, calculate the electric potential at the midpoint of the base.
V
Figure P16.17
1 answer