I need help with number 6 and 7 r-21_Electrostatics erickson (658710) 1n 6 LC 0

Question

I need help with number 6 and 7

r-21_Electrostatics erickson (658710) 1n 6 LC 0 1 1 234 5 6 7 8 9 Find the force on a charge of 6 C on the x axis at x = 7 cm. The value of the Coulomb constant is 8.98755 × 109 N·m2/C2 Answer in units of N 006 (part 1 of 2) 10.0 points A 5.4 C point charge is on the x-axis at 5 m , and a 4.9 C point charge is on the x-axis at 2.8 m Find the magnitude of the net electric field at the point on the y-axis where y = 2.3 m. The value of the Coulomb constant is 8.98755 × 109 N·m2/C2 Answer in units of N/C. 007 (part 2 of 2) 10.0 points Determine the direction of this electric field (as an angle between-180° and 180° mea- sured from the positive z-axis, with counter- clockwise positive.) Answer in units of° 008 10.0 points The electron gun in a television tube is used to accelerate electrons with mass 9.109 x 10-31 kg from rest to 3 x 107 m/s within a distance of 2.4 cm. What electric field is required? The funda- mental charge is 1.602 x 10-19 C.

Explanation / Answer

6)

q1 = 5.4 uC

(x1 , y1 ) = (5 , 0)

q2 = 4.9 uC

(x2 , y2) = (2.8 , 0)

point p is at ( x, Y) = (0 , 2.3)

electric field due to q1

distance of point p from q1 , r1 = sqrt((x - x1)^2 + (y-y1)^2)

r1 = sqrt(5^2+2.3^2) = 5.5 m

E1x = -k*q1*x1/r1^3 = -8.98755*10^9*5.4*10^-6*5/5.5^3 = -1458.5 N/C

E1y = k*q1*y/r1^3 = 8.98755*10^9*5.4*10^-6*2.3/5.5^3 = 671 N/C

electric field due to q2

distance of point p from q2 , r2 = sqrt((x - x2)^2 + (y-y2)^2)

r2 = sqrt(2.8^2+2.3^2) = 3.62 m

E2x = -k*q2*x2/r2^3 = -8.98755*10^9*4.9*10^-6*5/3.62^3 = -4641.7 N/C

E2y = k*q2*y/r2^3 = 8.98755*10^9*4.9*10^-6*2.3/3.62^3 = 2135.2 N/C

Ex = E1x + E2x = -1458.5 - 4641.7 = -6100.2 N/C

Ey = E1y + E2y = 671 + 2135.2 = 2806.2 N/C

E = sqrt(Ex^2 + Ey^2) = 6714.7 N <<<<------------ANSWER

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direction = tan^-1(Ey/Ex) = 155.3 degrees <<<<-----------ANSWER

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