EECE 380 Problems, Spring 2008

Electrostatics (cont.) Problem 1 Two positive charges, O and 4Q, are separated by a distance d. in air. A third charge is so placed that the entire system is in equilibrium. Determine the location, the magnitude, and the sign of the third charge. d io QO q 40 The unknown charge q has to be negative and located between O and 4Q, as shown in figure above. For equilibrium. F 12 = F32 = Fi3. Hence, Qq _ 40q _ (40XQ) ( d _ xy x? d 2 Which yield g = -4/90, x = 2/3d Problem 2 A2 uC charge is located at (0.3.0) m and a 4 uC charge is at (4.0.0) m. Determine E at (0.0.5) m. The electric field intensity of an isolated point charge at an arbitrary location is defined by R-R' Z E, ~_((R-R) (Vin) “ E 4e,|R-R} R, = R-R, = -3ay + Sa, R;=R-R, =-4a, + Sa, E 1 2*10° —3a, +5a, ~ 4x10” /367\34) 34 E, = -272.38a, + 453.97a, 4*10°% “a 2s E, = _ : 4n(l0* /367\41)\-V41 E, =-548.5la, +685.64a, E=E, +E, =~548.5la, —273.38a, +1139.6la, (Vim) Problem 3 A cube of 2-m side is centered at the origin with the edges parallel to the axes. Within the cube, D = (10x°/3)a, (C/m*). What is the total charge contained in the cube? p= VD =(10/3)(3x") = 10x” (C/m’) 0=[pdv =f [ 10x" dedyde = = Problem 4 Four point charges are located at the corners of a square, of length 2Z on each side, symmetrically located relative to the origin, as shown in the figure below. Derive an expression for the potential V at a position x along the axis. ' +O -O o«—__"——-e Vx) = 2.2.0.0) SR F 4me,\R, R, R, R, SAY Ro! 20(1 1 Mf V(x) =—=| —-— — 2L ©) Ane, \ R, z| ma > r Rye QO 1 1 wo RN 1(9)-52 | o . 2me,| Je +(L+xy Ve +(L-x) SF L ‘ ’ 2 <— 2L — 2 Problem 5 Given a line charge density p = 5 uC/m along the z axis, determine the potential difference between the points A(1m, 7.0) and B(3m, 7, 4m). The electric field intensity due to an infinite straight line charge of uniform density is given by: E=a,—"— (vm) 27€ 51 The electrostatic potential difference between B and A equals the work done in moving a unit charge from A to B, and is given by: V(B)-V(A)=-[, Eat B 35* 10% V(B)-V(A)=-| E,dr=- dr (B)-V(A)=-[ E,ai Ver 5*10° V(B)-V(A)=- In3 (V) 27€,