Charge Q is uniformly distributed over a line segment of length 2L, as shown bel
ID: 1530481 • Letter: C
Question
Charge Q is uniformly distributed over a line segment of length 2L, as shown below. At point P with co-ordinate (x, 0), the magnitude of the y-component of the electric field at point P is 0 1/4 pi element_0 Q/2squareroot L^2 + x^2 1/4 pi element_0 Q/squareroot L^2 + x^2 1/4 pi element_0 Q/x squareroot L^2 + x^2 1/4 pi element_0 Q/2x squareroot L^2 + x^2 g = 9.8 m/s^2 e = 1.6 times 10^-19 kg m_e = 9.1 times 10^-31 kg m_p = 1.66 times 10^-27 kg k = 9 times 10^9 N.m^2/C^2 = 1/4 pi element_0 Q = Ne rightarrow F = kq_1 q_2/r^2 r cap rightarrow E = lim_q_0 rightarrow 0 rightarrow F/q_0 rightarrow E = kq/r^2 r cap (distribution) rightarrow E = integral d rightarrow E = integral k dq/r^2 r capExplanation / Answer
the charge on this element is dQ=(Q*da)/(2L)
the electric field due to this element at 'p' is
dE= dQ/(4*pi*eo)*AP^2............
AP is distnce from small element and point P
the resultant field is allong prendicular bisector.....
the component dE along pependicular bisector is dE cos(theta)....
theta is the angle between perpendicular bisector and a line drawn from small element to point P
dE cos(theta) = x*Q*da/[4*pi*eo*2L(a^2+x^2)^(3/2)].....
net field at P is
E= integral [dE cos(theta)]....
by integrating we get
E= Q/[4*pi*eo*x(L^2+x^2)]