please solve question no 9
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we know potential due a point charge is given by
V = Kq/r
so
V is given = 10³V
and the distance = r = 0.3 m
=> V = K × q/0.3
=> V = 9×10^9 ×q/0.3
=> 10³ = 3×10^10 ×q
=> q = 1/3 × 10^-7
now we have to find the potential at the distance of 0.4 meter
so electric field intensity at the distance of 0.4 meter will be
E is given by Kq/r²
E = 9×10^9 × 1/3 × 10^-7/(0.4)²
E = 3/16 × 10⁴
E = 1875Vm-¹
ask me in comment section if you will face any problem
V = Kq/r
so
V is given = 10³V
and the distance = r = 0.3 m
=> V = K × q/0.3
=> V = 9×10^9 ×q/0.3
=> 10³ = 3×10^10 ×q
=> q = 1/3 × 10^-7
now we have to find the potential at the distance of 0.4 meter
so electric field intensity at the distance of 0.4 meter will be
E is given by Kq/r²
E = 9×10^9 × 1/3 × 10^-7/(0.4)²
E = 3/16 × 10⁴
E = 1875Vm-¹
ask me in comment section if you will face any problem
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