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# An electric field E=20i+30j exists in space.if the potential at the origin is taken to be zero,then what is the potential at (2m,2m)?

Arun
25763 Points
3 years ago
Dear student

Change in potential = - integral (vector E dot vector dl), dl is length element
So V-0 = -vector E dot integral(vector dl) from (0,0) to (2,2)

V= -(20i ̂+30j ̂ ) . {(2,2)-(0,0)}
V= -(20,30) . (2,2)
V= -(40+60)
V= -100V
Khimraj
3007 Points
3 years ago
potential change due to x component of E in y direction is zero as angle is 90 and similialy
potential change due to y component of E in x direction is zero
SO change in potential
V – 0 = – ( 20*2 + 30 *2) = -100V.
Keshav
15 Points
2 years ago
V=-ED
V=(20i+30j)*(2-0,2-0)
V=(20*2+30*2)
V=(-1)(100)
V=-100V
Change in potential = - integral (vector E dot vector dl), dl is length element
So V-0 = -vector E dot integral(vector dl) from (0,0) to (2,2)

Yash Chourasiya
one year ago
Hello Student

ΔV = −∫E.dl
where,dlisthelengthelement
V−0 = −E.∫dl {Limit from (0,0) to(2,2) }
V = −(20iˆ+30jˆ).{(2,2)−(0,0)}
V = −(20iˆ+30jˆ).(2iˆ+2jˆ)
V = −(40+60)
V = −100V