Some equipotential surface are shown in the figure. The magnitude and directions of the electric field is

$100 V / m$ making angle $120^{o}$ with the x-axis
None of the above
$200 V / m$ making angle $120^{o}$ with the x-axis
$100 V / m$ making angle $60^{o}$ with the x-axis

Question

Some equipotential surface are shown in the figure. The magnitude and directions of the electric field is

$100 V / m$ making angle $120^{o}$ with the x-axis
None of the above
$200 V / m$ making angle $120^{o}$ with the x-axis
$100 V / m$ making angle $60^{o}$ with the x-axis

Toppr Admin · Accepted Answer

Perpendicular distance between equipotential surfaces-d-10sin30o-5cmE-x394-Vd-30-x2212-205-100-200Vm-x2212-1Direction of electric field will be in the direction of decreasing potential

Some equipotential surface are shown in the figure. The magnitude and directions of the electric field is100V/m making angle 120o with the x-axisNone of the above200V/m making angle 120o with the x-axis100V/m making angle 60o with the x-axis

Perpendicular distance between equipotential surfaces:d=10sin30o=5cmE=ΔVd=30−205/100=200Vm−1Direction of electric field will be in the direction of decreasing potential

Some equipotential surface are shown in the figure. The magnitude and directions of the electric field is

$100 V / m$ making angle $120^{o}$ with the x-axis
None of the above
$200 V / m$ making angle $120^{o}$ with the x-axis
$100 V / m$ making angle $60^{o}$ with the x-axis

Perpendicular distance between equipotential surfaces:
$d = 10 s i n 30^{o} = 5 c m$
$E = \frac{Δ V}{d} = \frac{30 - 20}{5 / 100} = 200 V m^{- 1}$
Direction of electric field will be in the direction of decreasing potential