Question

A QL = -48 μC charge is placed on the x-axis at x = - 0.3 m. A QR = 48 μC charge is placed at x = +0.3 m. (for all answers below assume that the unit vector x^ points toward positive x, and y^ points towards positive y.).What is the x-component of the electric field at x = 0 m and y = 0.3 m?What is the x-component of the electric field at x = 0 m and y = -0.3 m? The answer is not -17X10^7 or -1.89X10^7

Answer 1

Take at the point A is qA = minus 48 mu and at B is qB = 48 mu c Then the electric field at y = plusminus 0.3m due to the qA is the alterative and due to the qB is repulsive now the magnitude of electric field y = plusminus 0.3m will be same due to both charge qA proptional to qB. and the magnitude of electric field at y = plusminus 0.3m by qA is FA = 1/4 pi e times qA/d^2 = = 2.4 times 10^6 And due to charge qB is FB = 1/4 pi e . QB/d^2 = g times 10^g times 48 times 10^6/(0.362 plus 0.3^2) = 2.4 times 10^6 Now, since , the two forces has same magnitude then the their y- components cancel out and only the x- component will survive, And value of x- component will be F= squareroot FA^2 PLUS FB^2 because FA^rightarrow and FB^rightarrow are perpendicular. = 3.394 times 10^6 n/e. a) At y = 0.3: x = 0. F^rightarrow times 10^6: n/e b) At y = 0.3m: x = 0 F^rightarrow times = minus 3.394 times 10^6: n/e