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Modeling Motion Electric Field Solutions 1 Two charges are placed in a line as shown below y m A 1 0 mC C 2 0 mC B x m a Find the electric field strength at points A B and C At A the electric field due to the 2 mC charge points right toward the negative charge and has magnitude E 9 109 2 10 3 42 1 125 106 N C and the electric field due to the 1 mC charge points left away from the positive charge and has magnitude E 9 109 1 10 3 22 2 25 106 N C So the total electric field is to the left with magnitude E 2 25 1 125 106 1 125 106 N C Similarly at B the electric field due to the 2 mC charge points left toward the negative charge and has magnitude E 9 109 2 10 3 22 4 5 106 N C and the electric field due to the 1 mC charge points right away from the positive charge and has magnitude E 9 109 1 10 3 42 0 5625 106 N C So the total electric field is to the left with magnitude E 4 5 5625 106 3 9 106 N C At C the magnitude of the electric field due to the 2 mC is E 9 109 2 10 3 5 3 6 106 N C pointing down toward the negative charge The x component is Ex E cos E 1 5 1 61 106 N C and the y component is Ey E sin E 2 5 3 22 106 N C Similarly the magnitude of the field due to the 1 mC charge at C is E 9 109 1 10 3 5 1 8 106 N C pointing up away from the negative charge The x component is Ex E cos E 1 5 0 805 106 N C and the y component is Ey E sin E 2 5 1 61 106 N C So the combined electric field has x component Ex 1 61 0 805 106 2 41 106 N C and the y component is Ey 3 22 1 61 106 1 61 106 N C b Find the location where the electric field is zero The electric field will be zero on the left of the 1 mC charge Here the fields due to each charge are opposite in direction and will be equal in magnitude at one point Let the point where the field is zero be d units to the left of the 1 mC charge Then the field due to the 1 mC charge has magnitude E k 1mC d2 and the field due to the 2 mC charge has magnitude E k 2mC 2 d 2 Setting these two equal implies that k d2 2k 2 d 2 2 d 2 2d2 d 2 2d d 2 2 1 4 83 m left of the 1 mC charge or at position x 5 83 c Repeat your calculation for the case where both charges are positive If both charges where positive then the fields would cancel between the two charges because it is here where the fields point in opposite directions Let the point where the field is zero be d units to the right of the 1 mC charge Then the electric field is zero when k d2 2k 2 d 2 2 d 2 2d2 2 d 2d d 2 2 1 0 83 m to the right of the 1 mC charge or at position x 0 17