Modeling Motion Electric Field Solutions1. Two charges are placed in a line as shown belowABC1.0 mC -2.0 mCx (m)y(m)(a) Find the electric field strength at points A, B and CAt A the electric field due to the -2 mC charge points right (toward the negative charge)and has magnitudeE = (9 × 109)(2 × 10−3)/42= 1.125 × 106N/Cand the electric field due to the 1 mC charge points left (away from the positive charge)and has magnitudeE = (9 × 109)(1 × 10−3)/22= 2.25 × 106N/C.So the total electric field is to the left with magnitudeE = (2.25 − 1.125 × 106= 1.125 × 106N/C.Similarly at B the electric field due to the -2 mC charge points left (toward the negativecharge) and has magnitudeE = (9 × 109)(2 × 10−3)/22= 4.5 × 106N/Cand the electric field due to the 1 mC charge points right (away from the positive charge)and has magnitudeE = (9 × 109)(1 × 10−3)/42= 0.5625 × 106N/C.So the total electric field is to the left with magnitudeE = (4.5 − .5625) × 106= 3.9 × 106N/C.At C the magnitude of the electric field due to the -2 mC isE = (9 × 109)(2 × 10−3)/5 = 3.6 × 106N/C,pointing down toward the negative charge. The x component isEx= E cos θ = E(1/√5) = 1.61 × 106N/Cand the y component isEy= −E sin θ = −E(2/√5) = −3.22 × 106N/C.Similarly the magnitude of the field due to the 1 mC charge at C isE = (9 × 109)(1 × 10−3)/5 = 1.8 × 106N/C,pointing up away from the negative charge. The x component isEx= E cos θ = E(1/√5) = 0.805 × 106N/Cand the y component isEy= E sin θ = E(2/√5) = 1.61 × 106N/C.So the combined electric field has x componentEx= (1.61 + 0.805) × 106= 2.41 × 106N/Cand the y component isEy= (−3.22 + 1.61) × 106= −1.61 × 106N/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 toeach charge are opposite in direction and will be equal in magnitude at one point. Letthe point where the field is zero be d units to the left of the 1 mC charge. Then thefield due to the 1 mC charge has magnitude E = k(1mC)/d2and the field due to the -2mC charge has magnitude E = k (2mC)/(2 + d)2. Setting these two equal implies thatk/d2= 2k/(2 + d)2⇒ (2 + d)2= 2d2⇒ d + 2 =√2d ⇒ d = 2/(√2 − 1) = 4.83 mleft 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 chargesbecause it is here where the fields point in opposite directions. Let the point where thefield is zero b e d units to the right of the 1 mC charge. Then the electric field is zerowhen k/d2= 2k/(2 −d)2⇒ (2 −d)2= 2d2⇒ 2 −d =√2d ⇒ d = 2/(√2 + 1) = 0.83m to the right of the 1 mC charge or at position x =