Answer:
Electric potential E = -45.631 KV
Explanation:
Given: q1 =3.80μC = 3.80 × 10∧-6 C, q2 =-7.10μC = -7.10 × 10∧-6 C
distance between them =1.30 m
at midpoint distance from q1 will be r1= 1.30 m r1= 0.65m and r2=0.65m
using Formula of Electric Potential E
E= [tex]\frac{K q1}{r1}[/tex] + [tex]\frac{K q2}{r2}[/tex]
E = [tex]\frac{K}{r}[/tex] (q1 + Q2) (∴ r= r1 = r2 )
E = 9.988×10^9 N.m²2/C² / .65 m ( 3.80 × 10^-6 C - 7.10 × 10^-6 C )
E = -45,631.3846 V
E = -45.631 KV
