On the X axis suppose you have two charges Q1 = -94C andis positioned at -38 cm

Question

On the X axis suppose you have two charges Q1 = -94C andis positioned at -38 cm on -x axis; Q2 = 127 C and ispositioned at 36cm on the + x axis: A) what is the electric potential at -5 cm on xaxis? B) Where could a charge be placed so that the electricpotential equals zero? C) If a 3rd charge Q3 = +25 C that weighs 19 gramsis released from the origin, what is its magnitude and directionwhen it is farthest away from Q1 and Q2? Please include actual given values in the answers. Thanks Not a text problem On the X axis suppose you have two charges Q1 = -94C andis positioned at -38 cm on -x axis; Q2 = 127 C and ispositioned at 36cm on the + x axis: A) what is the electric potential at -5 cm on xaxis? B) Where could a charge be placed so that the electricpotential equals zero? C) If a 3rd charge Q3 = +25 C that weighs 19 gramsis released from the origin, what is its magnitude and directionwhen it is farthest away from Q1 and Q2? Please include actual given values in the answers. Thanks Not a text problem

Explanation / Answer

two charges Q1 = -94C and is positioned at -38 cmon -x axis; Q2 = 127 C and is positioned at 36cm on the+ x axis A)the electric potential at -5 cm on x axis V = V1 + V2 = (1/4o) * (Q1/r1 +Q2/r2) (1/4o) = 9 * 10^9Nm2/C2 Q1 = -94 C = -94 * 10^-6C Q2 = 127 C = 127 * 10^-6 C Q1 = -94 C = -94 * 10^-6C Q2 = 127 C = 127 * 10^-6 C r1 = (-5 - (-38)) cm = 33 cm = 33 * 10^-2 m r2 = (36 - (-5)) cm = 41 cm = 41 * 10^-2 m B)third charge be placed at x = d from Q2 therefore (1/4o) * [(Q1/R1) + (Q2/R2)] = 0 or [(Q1/R1) + (Q2/R2)] = 0 or (Q1/R1) = -(Q2/R2) R1 = (74 + d) cm = (74 + d) * 10^-2 m R2 = d cm = d * 10^-2 m or (Q1/(74 + d) * 10^-2) = -(Q2/d * 10^-2) or (Q1/(74 + d) * 10^-2) = -(Q2/d * 10^-2) or d = -(74/(Q1/Q2 + 1)) -------------(1) negative value indicates the third charge should be placed inopposite direction. C)a 3rd charge Q3 = +25 C that weighs 19 grams isreleased from the origin therefore, (1/2)mv2 = Q3 * V V = (1/4o) * (Q3/d) (d from equation(1)) or (1/2)mv2 = Q3 * (1/4o) *(Q3/d) = (1/4o) * (Q3^2/d) or v = [(2/m) * (1/4o) *(1/d)]1/2 * Q3

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