A charge of -1.0 Solution solution for a different question of exactly same type

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A charge of -1.0

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solution for a different question of exactly same type with explanaton .. so if it helps please rate me first A charge q1 = 5 µC, is at the origin, and a second charge, q2 = -3 µC, is on the x-axis 1 m from the origin. Find the electric field at a point on the y-axis 0.5 m from the origin. Magnitude of Field: Direction of Field [ angle between 0 and 359 °] draw charge 1 and charge 2 on the x-y axis... from this, you can get get the vector length between charge 2 on the x-axis and the point 0.5 on the y-axis, as well as the angle. sqrt(0.5²+1²) = 1.12 m ---> vector length from charge 2 to desired point theta = atan(0.5/1) = 26.6º ---> angle above x axis for that vector the electric field is a vector quantity, so E1 + E2 = E E = F / q , where F = k q1 q2 / r² for E1 (all in the j direction) E1 = k q1 q2 / (q2 r²) = kq1 / r² j = (8.99E9)(5E-6)/(0.5²) = 179800 N/C j for E2 (i and j directions) E2 = k q1 q2 / (q1 r²) = kq2 / r² (cos ? i + sin ? j) = [(8.99E9)(-3E-6) / (1.12²)] (cos 26.6º i + sin 26.6º j) = - 19225 N/C i - 9627 N/C j Now add E1 + E2 E = -19255 i + ( 179800 - 9627) j = -19255 i + 170173 j E = sqrt(19255² + 170173² ) E = 171259 N/C

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