(a) Find the electric potential, taking zero at infinity, at theupper right corn
ID: 1762407 • Letter: #
Question
(a) Find the electric potential, taking zero at infinity, at theupper right corner (the corner without a charge) of the rectanglein the figure below. (y = 3.8cm and x = 6.3 cm)1Answer in J/C
(b) Repeat if the 2.00 µC charge is replaced with a charge of-2.00 µC.
2Answer in J/C
X
8.0 (mico C)----------------
l l
l l
ly l
l l
l...........................................4.0 (micro C)
2.0 (micro C)
Explanation / Answer
The electric potential due to the charge 2.0 C at theupper right corner is V1 = k * (q1/r1) k = (1/4o) = 9 * 109Nm2/C2,q1 = 2.0 C = 2.0 *10-6 C and r1 = [(6.3)2 +(3.8)2]1/2 = 7.3 cm = 7.3 * 10-2m The electric potential due to the charge 4.0 C at theupper right corner is V2 = k * (q2/r2) q2 = 4.0 C = 4.0 * 10-6 C andr2 = 3.8 cm = 3.8 * 10-2 m The electric potential due to the charge 8.0 C at theupper right corner is V3 = k * (q3/r3) q3 = 8.0 C = 8.0 * 10-6 C andr3 = 6.3 cm = 6.3 * 10-2 m The total electric potential at the upper right corneris V = V1 + V2 + V3 = k *(q1/r1) + k * (q2/r2) +k * (q3/r3) = k *((q1/r1) + (q2/r2)+ (q3/r3)) (b)When the 2.00 µC charge is replaced with a charge of-2.00 µC then the electric potential V2becomes V2' = k * (q2'/r2) where q2' = -2.00 C = -2.00 * 10-6C The total electric potential at the upper right corneris V = V1 + V2' + V3 = k *(q1/r1) + k * (q2'/r2)+ k * (q3/r3) = k *((q1/r1) +(q2'/r2) +(q3/r3)) The electric potential due to the charge 4.0 C at theupper right corner is V2 = k * (q2/r2) q2 = 4.0 C = 4.0 * 10-6 C andr2 = 3.8 cm = 3.8 * 10-2 m The electric potential due to the charge 8.0 C at theupper right corner is V3 = k * (q3/r3) q3 = 8.0 C = 8.0 * 10-6 C andr3 = 6.3 cm = 6.3 * 10-2 m The total electric potential at the upper right corneris V = V1 + V2 + V3 = k *(q1/r1) + k * (q2/r2) +k * (q3/r3) = k *((q1/r1) + (q2/r2)+ (q3/r3)) (b)When the 2.00 µC charge is replaced with a charge of-2.00 µC then the electric potential V2becomes V2' = k * (q2'/r2) where q2' = -2.00 C = -2.00 * 10-6C The total electric potential at the upper right corneris V = V1 + V2' + V3 = k *(q1/r1) + k * (q2'/r2)+ k * (q3/r3) = k *((q1/r1) +(q2'/r2) +(q3/r3)) The electric potential due to the charge 8.0 C at theupper right corner is V3 = k * (q3/r3) q3 = 8.0 C = 8.0 * 10-6 C andr3 = 6.3 cm = 6.3 * 10-2 m The total electric potential at the upper right corneris V = V1 + V2 + V3 = k *(q1/r1) + k * (q2/r2) +k * (q3/r3) = k *((q1/r1) + (q2/r2)+ (q3/r3)) (b)When the 2.00 µC charge is replaced with a charge of-2.00 µC then the electric potential V2becomes V2' = k * (q2'/r2) where q2' = -2.00 C = -2.00 * 10-6C The total electric potential at the upper right corneris V = V1 + V2' + V3 = k *(q1/r1) + k * (q2'/r2)+ k * (q3/r3) = k *((q1/r1) +(q2'/r2) +(q3/r3)) The electric potential due to the charge 8.0 C at theupper right corner is V3 = k * (q3/r3) q3 = 8.0 C = 8.0 * 10-6 C andr3 = 6.3 cm = 6.3 * 10-2 m The total electric potential at the upper right corneris V = V1 + V2 + V3 = k *(q1/r1) + k * (q2/r2) +k * (q3/r3) = k *((q1/r1) + (q2/r2)+ (q3/r3)) (b)When the 2.00 µC charge is replaced with a charge of-2.00 µC then the electric potential V2becomes V2' = k * (q2'/r2) where q2' = -2.00 C = -2.00 * 10-6C The total electric potential at the upper right corneris V = V1 + V2' + V3 = k *(q1/r1) + k * (q2'/r2)+ k * (q3/r3) = k *((q1/r1) +(q2'/r2) +(q3/r3)) The total electric potential at the upper right corneris V = V1 + V2' + V3 = k *(q1/r1) + k * (q2'/r2)+ k * (q3/r3) = k *((q1/r1) +(q2'/r2) +(q3/r3))Related Questions
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