Calculate the electric field half way between two charges, where one charge is +

Question

Calculate the electric field half way between two charges, where one charge is + 21.0 nC and the second charge is 4.5 nC at a distance of 239 mm. (Note that the second charge may be given as positive or negative, read the value carefully.)

Express the result in [N/C], and to three significant figures. If you must use scientific notation, please enter as follows: ex. 0.00012 = 1.2E-4. Only answer in numerical values, no units. For negative numbers, leave no space between the negative sign and the number (right: -1.00, wrong, - 1.00).

Explanation / Answer

q1 = 21*10^-9 C q2 = 4.5*10^-9 C Distance between two charges d = 239 mm = 0.239 m Let P be the point between the two charges. Seperation between point P and charge, r = d/2 = 0.1195 m Electric field at P due to q1, E1 = Kq1/r^2                                                 = (9*10^9 * 21*10^-9)/(0.1195)^2                                                 = 13235.06( i) N/C Electric field due to q2 at P, E2 = Kq2/r^2                                                 = (9*10^9 * 4.5*10^-9)/(0.1195)^2                                                 = 2836.08 ( - i ) N/C Since both charges are +ve, E1 and E2 are in opposite directions to each other.So resultant electric field                 E = E1 + E2                            = 13235.06(i) - 2836.08(i)                            = 10398.98 i N/C                            = 1.04E4 N/C    Since both charges are +ve, E1 and E2 are in opposite directions to each other.So resultant electric field                 E = E1 + E2                            = 13235.06(i) - 2836.08(i)                            = 10398.98 i N/C                            = 1.04E4 N/C   

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