a uniform electric field of side 60 volt per metre exist in xy plane making an angle 120 degree with positive x-axis the potential difference between point p(-2,3) and q(4,1) is given by
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Steps and Calculation :
1) E = -60 cos (60°) i. + 60 sin 60° j
E = -30i +30√3 j
R(PQ) = (4-(-2))i + (1-3)j
R(PQ) = 6i -2 j
2) We know that,
V(PQ) = -[E. R(PQ)]
=> V(PQ) = -[(-30i + 30√3j) . (6i -2j)]
=> V(PQ) = 30* 6 + 30 * 2√3
=> V(PQ) = 180 + 60√3 = 60√3(√3+ 1)V
Hence,
Magnitude of Potential difference between P and Q is 60√3(√3+1) V.
1) E = -60 cos (60°) i. + 60 sin 60° j
E = -30i +30√3 j
R(PQ) = (4-(-2))i + (1-3)j
R(PQ) = 6i -2 j
2) We know that,
V(PQ) = -[E. R(PQ)]
=> V(PQ) = -[(-30i + 30√3j) . (6i -2j)]
=> V(PQ) = 30* 6 + 30 * 2√3
=> V(PQ) = 180 + 60√3 = 60√3(√3+ 1)V
Hence,
Magnitude of Potential difference between P and Q is 60√3(√3+1) V.
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