Name: Electric and Gravitational Forces
Uploaded: 2019-07-19
Duration: 3 min 11 s
Description: Do you know the difference and similarities between the gravitational force and electrostatic force? Watch this video to explore it. Don't miss it.

Question

Four charges are placed at the vertices of a square of side $5.00 cm$ , as shown in the diagram.

On a copy of the diagram, show the forces acting on $2.0 μC$ charge. Find the magnitude and the direction of the net force on the $2.0 μC$ charge. (use $K = 9 \times 10^{9}$ .)

Accepted Answer

The below figure shows the force acting $2.0 μC$ charge.

From the figure, the distance between the $4 μC$ and the $2 μC$ is equal to the length of the diagonal of the square.

By the Pythagorean theorem, the length of the diagonal is

$d = \sqrt{5^{2} + 5^{2}} = 5 \sqrt{2} cm$

The distance between the other two charges is $5 cm$ .

The force due to each charge on the $2 μC$ is

$F = \frac{k q_{1} q_{2}}{r^{2}}$

here, $r$ is the distance between the two charges.

$F_{1} = \frac{9 \times 10^{9} \times 2 \times 10^{- 6} \times 3 \times 10^{- 6}}{{(5 \times 10^{- 2})}^{2}} = 21.6 N$

$F_{2} = \frac{9 \times 10^{9} \times 2 \times 10^{- 6} \times 4 \times 10^{- 6}}{{(5 \sqrt{2} \times 10^{- 2})}^{2}} = 14.4 N$

$F_{3} = \frac{9 \times 10^{9} \times 2 \times 10^{- 6} \times 1 \times 10^{- 6}}{{(5 \times 10^{- 2})}^{2}} = 7.2 N$

Let $F_{2}$ makes angle $θ$ from the horizontal

$\tan (θ) = \frac{5}{5} = 1$

$\Rightarrow θ = 45 °$

So, the horizontal and vertical components of the $F_{2}$ vector are

$F_{2 H} = F_{2} \cos (45 °) = 14.4 \times \frac{1}{\sqrt{2}} = 10.2 N$

$F_{2 V} = F_{2} \sin (45 °) = 14.4 \times \frac{1}{\sqrt{2}} = 10.2 N$

So, from the figure the net force in the horizontal and vertical direction is

$F_{H} = F_{2 H} - F_{3} = 10.2 - 7.2 = 3 N$

$F_{V} = F_{2 V} - F_{1} = 10.2 - 21.6 = - 11.4 N$

The net force on the charge $F_{n e t} = \sqrt{{(F_{H})}^{2} + {(F_{V})}^{2}}$

$F_{n e t} = \sqrt{3^{2} + {(- 11.4)}^{2}} = 11.8 N$

The direction of the force is

$ϕ = \tan^{- 1} \frac{F_{V}}{F_{H}} = \tan^{- 1} (\frac{- 11.4}{3}) = - 75.3 °$

Four charges are placed at the vertices of a square of side $5.00 cm$ , as shown in the diagram.

On a copy of the diagram, show the forces acting on $2.0 μC$ charge. Find the magnitude and the direction of the net force on the $2.0 μC$ charge. (use $K = 9 \times 10^{9}$ .)

Important Questions on Fields (HL)

Learn and Prepare for any exam you want !

Four charges are placed at the vertices of a square of side 5.00 cm, as shown in the diagram. On a copy of the diagram, show the forces acting on 2.0 μC charge. Find the magnitude and the direction of the net force on the 2.0 μC charge. (use K=9×109.)

Important Questions on Fields (HL)

Learn and Prepare for any exam you want !

Four charges are placed at the vertices of a square of side $5.00 cm$ , as shown in the diagram.

On a copy of the diagram, show the forces acting on $2.0 μC$ charge. Find the magnitude and the direction of the net force on the $2.0 μC$ charge. (use $K = 9 \times 10^{9}$ .)