Two positively charged particles each having charge 20 C, are kept at a distance
of 2.0 m from each other. Determine the force on each charge and the electric field
due to each charge. Show the force and electric field vectors on appropriate
diagrams. What is the resultant force at a point midway from the two charges along
the straight line joining them?
Answers
Answer:
nations & Calculations
The electric field intensity is defined as the force imposed on a unit positive charge due to the other surrounding electric charges.
It is a vector quantity then & during a drawing of field intensity, that should be considered.
By Coulomb's law, the magnitude of the intensity due to an isolated charge is given by
\qquad\qquad \begin{aligned} \small |\bold{\overrightarrow E}|&= \small \frac{1}{4\pi \epsilon}\frac{|Q|}{r^2} \end{aligned}
∣
E
∣
=
4πϵ
1
r
2
∣Q∣
direction is in line with the positive unit charge
Therefore, from a single of the given charges, the generated field magnitude at an arbitrary point is
\qquad\qquad \begin{aligned} \small |\bold{\overrightarrow E_1}|&= \small \frac{1}{4\pi \epsilon}\frac{(+20nC)}{r_1} \end{aligned}
∣
E
1
∣
=
4πϵ
1
r
1
(+20nC)
When both the charges are considered,
\qquad\qquad \begin{aligned} \small |\overrightarrow E_{net}|&= \small |\overrightarrow E_1|+|\overrightarrow E_2|\\ &= \small \frac{1}{4\pi \epsilon}\frac{20}{r_1}+\frac{1}{4\pi \epsilon}\frac{20}{r_2}\\ &= \small \frac{20}{4\pi \epsilon}\Big[\frac{1}{r_1}-\frac{1}{r_2}\Big] \end{aligned}
∣
E
net
∣
=∣
E
1
∣+∣
E
2
∣
=
4πϵ
1
r
1
20
+
4πϵ
1
r
2
20
=
4πϵ
20
[
r
1
1
−
r
2
1
]
And it is the vector addition which gives the direction of the field at that point.
Given info : Two positively charged particles having charge 20μC are kept at a distance of 2m from each other.
To find :
- the force on charge.
- electric field due to each charge
- show the force and electric field vector on appropriate diagram.
- the resultant electric field at a point midway from the two charge along the straight line joining them.
solution : from Coulomb's law,
electric field due to each charge, E = kq/r²
= (9 × 10^9 × 20 × 10^-6)/(2)²
= 45 × 10³ N/C
= 4.5 × 10⁴ N/C
electric force due to each charge , F = qE
= 20 × 10^-6C × 4.5 × 10⁴ N/C
= 90 × 10¯² N
= 0.9 N
we know, two same charge oppose each other. so the electric force due to first charge to 2nd must be repulsive and vice-versa. and the direction of electric field at the point away from it. see figure.
the resultant electric at a point midway , F' = kq/(r/2)² - kq/(r/2)²
= 0
