Answer with Explanation:
a. Option d is true.
a negatively charged plane parallel to the end faces of the cylinder
b. Radius of cylinder, r=0.66m
Magnitude of electric field, E=300 N/C
We have to find the net flux through the closed surface.
Net electric flux,
c.
Net charge,
Where
Where
assuming north-south along the Y-axis and east-west along the X-axis
X = total X-displacement
from the graph , total displacement along X-direction is given as
X = 0 - 20 + 60 Cos45 + 0
X = 42.42 - 20
X = 22.42 m
Y = total Y- displacement
from the graph , total displacement along Y-direction is given as
Y = 40 + 0 + 60 Sin45 + 50
Y = 90 + 42.42
Y = 132.42 m
magnitude of the net displacement vector using Pythagorean theorem is given as
magnitude : Sqrt(X² + Y²) = Sqrt(22.42² + 132.42²) = 134.31 m
Direction : tan⁻¹(Y/X) = tan⁻¹(132.42/22.42) = 80.4 deg north of east
