Ch23 Homework Solutions

23.***.A charge of - 4.0 microCoulombs is located at the origin and a charge of - 5.0 microCoulombs is located along the y-axis at y = 2.0 m. At what point along the y-axis is the electric field zero?E = k q/r

^{2}r

_{1}= yr

_{2}= (2.0 m - y)r

_{2}^{2}= (2.0 m - y)^{2}r

_{2}^{2}= 4 - 4 y + y^{2}where y must now be measured in meters.

For point P

betweenthe two charges, 0 < y < 2.0 m,E_{1}pointsdownandE_{2}pointsupso the total field isE_{tot}=E_{net}=E_{2}-E_{1 .}

E_{tot}=E_{net}=E_{2}-E_{1 }= 0

E_{tot}= k q_{2}/r_{2}^{2}- k q_{1}/r_{1}^{2 }We have already taken care of the direction of the electric field due to the signs of the charges and the location of point P. In the following equations the q's are just the magnitudes.

E_{tot}= k (5 C)/(4 - 4 y + y^{2}) - k (4 C)/y^{2 = 0 }5 /(4 - 4 y + y

^{2}) - 4 /y^{2 = 0}5 y

^{2}/[y^{2}(4 - 4 y + y^{2})] - 4 (4 - 4 y + y^{2}) /^{[y2 (4 - 4 y + y2)] = 0}[5 y

^{2}- 4 (4 - 4 y + y^{2})] /^{[y2 (4 - 4 y + y2)] = 0}[5 y

^{2}- 4 (4 - 4 y + y^{2})]^{ = 0}5 y

^{2}- 4 (4 - 4 y + y^{2})^{ = 0}5 y

^{2}- 16 + 16 y - 4 y^{2}^{ = 0}y

^{2}- 16 + 16 y^{= 0}y

^{2}+ 16 y - 16^{= 0}We can now solve this using the

quadradic equation; the solutions arey = - 8.06 m

ory = + 0.94 mOf course, the

PHYSICSof the situation requires thaty = 0.94 m

Return to Homework Set(c) Doug Davis, 2002; all rights reserved