Question

Four point charges, each of magnitude 9.87 µC, are placed at the corners of a square 36.7 cm on a side. If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge. The value of Coulomb’s constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of N.

Answer 1

Answer:

Explanation:

charge on each corner, q = 9.87 micro coulomb

Side of square, a = 36.7 cm

Coulombic constant, K = 8.98 x 10^9 Nm²/C²

Ab = BC = CD = DA = 36.7 cm = 0.367 m

Diagonal, AC = BD = 1.414 x 0.367 = 0.52 m

Electric field at D due to charge at A

$E_{A}=\frac{Kq}{AD^{2}}$

$E_{A}=\frac{8.98\times 10^{9}\times 9.87\times 10^{-6}}{0.367^{2}}$

EA = 658053.74 N/C  

Electric field at D due to charge at C

$E_{C}=\frac{Kq}{CD^{2}}$

$E_{C}=\frac{8.98\times 10^{9}\times 9.87\times 10^{-6}}{0.367^{2}}$

EC = 658053.74 N/C  

Electric field at D due to charge at B

$E_{B}=\frac{Kq}{BD^{2}}$

$E_{B}=\frac{8.98\times 10^{9}\times 9.87\times 10^{-6}}{0.52^{2}}$

EB = 327783.28 N/C  

Resolve the compoents

Ex = EA + EB cos 45

Ex = 658053.74 + 327783.28 x 0.707

Ex = 889831.52 N/C

Ey = EC + EB Sin 45

Ey = 658053.74 + 327783.28 x 0.707

Ey = 889831.52 N/C

The resultant electric field is

$E = \sqrt{E_{x}^{2}+E_{y}^{2}}$

E = 1.414 x 889831.52 = 1258221.77 N/C

the electric force on the negative charge is

F = q x E

F = 9.87 x 10^-6 x 1258221.77

F = 12.42 N