Question

A 90 μC point charge is at the origin.

(a) Find the electric field at the point x1= 45 cm , y1= 0
(b) Find the electric field at the point x3= -20 cm , y3 = 60 cm.

Answer 1

Answer:

a) $E_1=4\times 10^6\ N.C^{-1}$

b) $E_3=202.5\ N.C^{-1}$

Explanation:

Given:

charge at the origin, $q=90\times 10^{-6}\ C$

a)

Electric field at point $(45\ cm,0\ cm)$:

The distance form the charge:

$d_1=0.45\ m$

Now the electric field at the given point:

$E_1=\frac{1}{4\pi.\epsilon_0} \times \frac{q}{d_1^2}$

$E_1=9\times 10^9\times \frac{90\times 10^{-6}}{0.45^2}$

$E_1=4\times 10^6\ N.C^{-1}$

b)

Electric field at point $(-20\ cm, 60\ cm)$:

The distance form the charge:

$d_3=\sqrt{(-20-0)^2+(60-0)^2}$

$d_3=63.2456\ m$

Now the electric field at the given point:

$E_3=\frac{1}{4\pi.\epsilon_0} \times \frac{q}{d_3^2}$

$E_3=9\times 10^9\times \frac{90\times 10^{-6}}{4000}$

$E_3=202.5\ N.C^{-1}$

Answer 2

(a) The required value of electric field at point x1 = 45 cm, y1 = 0 cm is $4 \times 10^{6} \;\rm N/C$.

(b) The required magnitude of electric field at point x3 and y3 is 202.5 N/C.

Given data:

The magnitude of point charge is, $q = 90 \;\rm \mu C =90 \times 10^{-6} \;\rm C$.

The problem is based on the concept of the electric field. The region where the electric force is experienced by the charged entities is known as the electric field.

(a)

The net distance at x1 = 45 cm = 0.45 m  and y1 = 0 m is,

$d=\sqrt{0.45^{2}+0^{2}}\\\\ d = 0.45 \;\rm m$

Then the expression for the net electric field is,

$E = \dfrac{k \times q}{d^{2}}$

here, k is Coulomb's constant.

Solving as,

 $E = \dfrac{9 \times 10^{9} \times 90 \times 10^{-6}}{0.45^{2}}\\\\ E = 4 \times 10^{6} \;\rm N/C$

Thus, the required value of electric field at point x1 = 45 cm, y1 = 0 cm is $4 \times 10^{6} \;\rm N/C$.

(b)

Similarly, the net distance at the point x3 = -20 cm = -0.2 m and y3 = 60 cm = 0.6 m is,

$d' = \sqrt{(-20)^{2}+60^{2}}\\\\ d' = 63.24 \;\rm m$

Also, the expression for the strength of electric field at point x3 and y3 is,

$E' = \dfrac{k \times q}{d'^{2}}$

Solving as,

$E' = \dfrac{9 \times 10^{9} \times 90 \times 10^{-6}}{63.24^{2}}\\\\ E' = 202.5 \;\rm N/C$

Thus, the required magnitude of electric field at point x3 and y3 is 202.5 N/C.

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