Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

What is the magnitude of the electric field at a point 0.0055 m from a 0.0025
C charge?
kg
Use E = and k=9.00 x 10 N.m²/C2.
O A. 7.4 x 1011 N
O B. 2.0 x 1010 N
O C. 4.1 x 10°N
OD. 7.9 x 1012 N



Sagot :

Answer:

Explanation:

The equation for the electric field is

[tex]E=\frac{kQ}{r^2}[/tex] so filling in:

[tex]E=\frac{9.00*10^9(.0025)}{(.0055)^2}[/tex] which in the end gives you

E = 7.4 × 10¹¹, choice A