Question

9, E 5305.79 N/C 2.-i 10, E = 7046.76 N/C 3. Question 6, chap 123, sect 7 part 1 of 2 10 points Three point charges are placed at the ver- - tices of an equilateral triangle. 5.-2i+j) -9 , 을 (2i+j) Find the magnitude of the electric field vec-9. 5 tor |Ell at P 10. 26- 5 kq 2 kq 1 kq 2 k Question 8, chap 123, sect 10 A rod 5.6 cm long is uniformly charged and The Determine the magnitude of the electric part 1 of 10 points has a total charge of-16.4 μC. Coulomb constant is 8.98755 x 10 N m2/C2 field along the axis of the rod at a point 47.7494 cm from the center of the rod. Answer in units of N/C. 2 kq 6. El Question 9, chap 123, sect 12 part 1 of 410 points 4 kq 4 kq 3 kq 1 kq A uniformly charged ring of radius 7.1 cnm has a total charge of 43 ,0. The value of the Coulomb constant is 8.98755 x 10 N m2/C2 Find the magnitude of the electric field on the axis of the ring at 1.15 em from the center of the ring. Answer in units of N/C. Question 7, chap 123, sect 7 part 2 of 2 10 points Question 10, chap 123, sect 12. part 2 of 4 10 points Find the direction of the field vector E at Find the magnitude of the electric field on
Please show step by step solution

Answer 1

As HOMEWORKLIB RULES guide lines i did first two only(first problem )

6) From figure ,

the electric field due to a negative point charge is away from the point .So,bottom two charges in the equilateral triangle produces electric field equal in magnitude and opposite in direction in the horizontal direction .Hence,the net horizontal electric field is zero. the net electric field is only vertical component of electric field at point p is due to top negative charge in the triangle.

electric field due to point charge E =k*q/(r^2)

the distance between top charge and point p =sqrt(a^2 - (a/2)^2)

r = sqrt(3*(a^2/4))

Ep =k*q/((sqrt(3*(a^2/4)))^2)

Ep =(4/3)*(k*q)/(a^2)

option (7) is correct

7)

the direction of electric field at point is upward. i.e (j) direction.

option (3) is correct