Question

Question 2 A charged sphere has a radius of 26.9 cm. The electric field strength at a distance 34.0 cm is 1.20x103 N/C. If the sphere's charge is halved, what would the strength of the electric field at 34.0 cm become? 2.40E+03 N/C 600. N/C 1.20E+03 N/C 300. N/C Question 3 Two charged, square plates, separated by a distance of 5.40 cm and each with a side length of 18.1 cm, make up a parallel-plate capacitor. The electric field inside the plates is 1.10x103 N/C. If the charge is doubled, what would the electric field inside the plates become? 275 N/C 1.10E+03 N/C 550. N/C 2.20E+03 N/C

Answer 1

2] Electric field E = kq/r^2

so if q is halved, E will be halved too.

new E = 600 N/C

second option

3] Electric field between capacitor = sigma/e0

= if charge will be doubled, sigma will be doubled, Electric field will also be doubled,

2.20E3 N/C

fourth option is correct