Question

Two point charges, +3.80 μC and -7.10 μC, are separated by 1.30 m. What is the electric potential midway between them?

Answer 1

Electric potential E = -45.631 KV

Step-by-step 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=
`(K q1)/(r1)` +
`(K q2)/(r2)`

E =
`(K)/(r)` (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