Question

Three negative point charges lie along a line as shown in (Figure 1). Point P lies 6.00 cm from the q2 charge measured perpendicular to the line connecting the three charges. Assume that q1 = -5.05 μC and q2  = -2.45 μC. 

Part A 

Find the magnitude of the electric field this combination of charges produces at point P

Part B 

Find the direction of the electric field this combination of charges produces at point P. 

Answer 1

here,

a = 6 cm = 0.06 m

b = 8 cm = 0.08 m

theta = arctan(b/a) = 53.13 degree

q1 = - 5.05 uC = - 5.05 * 10^-6 C

q2 = - 2.45 uC = - 2.45 * 10^-6 C

the electric field at point P , E = K * q1 /(a^2 + b^2) * ( - cos(theta) i + sin(theta) j) + K * q1 /(a^2 + b^2) * ( - cos(theta) i - sin(theta) j) - K * q2 /a^2 i

E = 9 * 10^9 * 10^-6 * (- 2 * 5.05 * cos(53.13) /(0.06^2 + 0.08^2) i - 2.45 /0.06^2 i) N/C

E = - 1.16 * 10^7 i N/C

a)

the magnitude of electric field is 1.16 * 10^7 N/C

b)

the direction of (towards charge q2)