A charge of 5 Solution draw charge 1 and charge 2 on the x-y axis... from this,

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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 0.02 on the y-axis, as well as the angle. sqrt(0.02²+0.02²) = 0.028 m ---> vector length from charge 2 to desired point theta = atan(0.02/0.02) = 45º ---> 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)/(0.02²) = 112500* 10^3 N/C j for E2 (i and j directions) E2 = k q1 q2 / (q1 r²) = kq2 / r² (cos ? i + sin ? j) = [(8.99E9)(-5E-6) / (0.028²)] (cos 45º i + sin 45 j) = - 40586*10^3 N/C i - 40586 * 10^3 N/C j Now add E1 + E2 E = -40586 i + ( 112500 - 40586) j = -40586 i + 71644 j * 10^3

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