a circular disc of radius 1/2√π m is placed in electric field 2i-3j+8k V/m in xy plane then electric flux through the disc will be
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see figure, a circular disc is placed in xy plane then, area vector is normal to xy plane e.g., along z - direction.
so, area vector of circular disc, A = πR² k m²
= π(1/2√π)² k m² [ given, R = 1/2√π m]
= π/4 k m²
given, electric field , E = 2i - 3j + 8k V/m
we know, electric flux = dot product of electric field and area vector.
= E.A
= (2i - 3j + 8k) . (π/4 k)
= 8 × π/4 = 2π Vm
hence, electric flux through the disc will be 2π Vm.
so, area vector of circular disc, A = πR² k m²
= π(1/2√π)² k m² [ given, R = 1/2√π m]
= π/4 k m²
given, electric field , E = 2i - 3j + 8k V/m
we know, electric flux = dot product of electric field and area vector.
= E.A
= (2i - 3j + 8k) . (π/4 k)
= 8 × π/4 = 2π Vm
hence, electric flux through the disc will be 2π Vm.
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