hysics 262 Spring 2018 Exam 1 Version: E Please solve the problem on the attache

Question

hysics 262 Spring 2018 Exam 1 Version: E Please solve the problem on the attached long answer template. Starting from fundamental formulas and showing all work. 1) Two point charges, Q1 =-4.0pC and Q2 = + 7.0 C, are placed as shown in the figure. (k = 1/4m0-8.998 109 N-m2/C2) What is the y component of the electric field,at the origin O? 19 m 1.2m 1Q1 an 2) Two point charges of +50Cand-9.0 C are located on the x-ax is at x-10 cnn and-+2.0 cm respectively. Where should a third charge of +3.0-uC be placed on the +x-axis so that the potential the origin is equal to zero? (14= 8.998 109 N·m2/C2) 5

Explanation / Answer

1) due to q1, field is along x axis hence y-component = 0

due to q2:

E = - k Q2 / (1.2^2 + 1.9^2)

Ey = - Ey (1.9)/sqrt(1.2^2 + 1.9^2)

Ey = - (9 x 10^9 x 7 x 10^-6 x 1.9) / (1.9^2 + 1.2^2)^(3/2)

Ey = - 10548 N/C

2) V = k Q /r

so potential at origin,

0 = (9 x 10^9) [ (5 x 10^-6 / 1) + (-9 x 10^-6 / 2) + (3 x 10^-6 / x)]

5 - 4.5 + (3/x) = 0

x = - 6 cm  

x is the distance from origin, that can not be negative,

that means there is no position where we can put third charge such that potential becomes zero at origin.

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