With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees

ELECTRIC FORCE (F)

F = $\frac{KQq}{d^{2} }$

Where K = 9 x $10^{9}$ N $m^{2}$ / $C^{2}$

The distance between $q_{1}$ and $q_{3}$ can be calculated by using Pythagoras theorem.

d = $\sqrt{33^{2} + 33^{2} }$

d = 46.7 cm = 0.467 m

For force $F_{1}$ , substitute all the parameters into the formula above

$F_{1}$ = (9 x $10^{9}$ x 3 x 1)/ $0.467^{2}$

$F_{1}$ = 2.7 x $10^{10}$ /0.218

$F_{1}$ = 1.24 x $10^{11}$ N

For force $F_{4}$ , substitute all the parameters into the formula above

$F_{4}$ = (9 x $10^{9}$ x 3 x 4)/ $0.33^{2}$

$F_{4}$ = 1.08 x $10^{11}$ /0.1089

$F_{4}$ = 9.92 x $10^{11}$ N

For force $F_{2}$ , substitute all the parameters into the formula above

$F_{2}$ = (9 x $10^{9}$ x 3 x 2)/ $0.33^{2}$

$F_{2}$ = 5.4 x $10^{10}$ /0.1089

$F_{2}$ = 4.96 x $10^{11}$ N

Summation of forces on Y component will be

$F_{y}$ = $F_{4}$ - $F_{1}$ Sin 45

$F_{y}$ = 9.92 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{y}$ = 9.04 x $10^{11}$ N

Summation of forces on X component will be

$F_{x}$ = $F_{2}$ - $F_{1}$ Cos 45

$F_{x}$ = 4.96 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{x}$ = 4.08 x $10^{11}$ N

Net Force = $\sqrt{F_{x} ^{2} + F_{y} ^{2} } }$

Net force = $\sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2} }$

Net force = 9.9 x $10^{11}$ N

The direction will be

Tan ∅ = $F_{y}$ / $F_{x}$

Tan ∅ = 9.04 x $10^{11}$ / 4.08 x $10^{11}$

Tan ∅ = 2.216

∅ = $Tan^{-1}$ (2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

8 0

1 year ago

Read 2 more answers