Two tiny spheres carrying charges of 1 microC and 3microC are placed 8 cm apart in air. What is the coulomb force on a unit positive test charge placed at the mid-point of the joining the two charges?
(1)9/16×10^7 N
(2)9/64×10^5 N
(3)9/4×10^5 N
(4)9/8×10^7 N
plzzz answer my question with explanation and if you don't know the correct solution plzz leave the question as it is....and don't take the advantage of points by writing rubbish like some of you have done in my previous questions.....plz understand what I'm saying...it's urgent so answer this with detailed explanation....if you will give me the correct answer I'll follow you and mark you as brainlist
Answers
Answer:
Here q
1
=1.8μC=1.8×10
−6
C,q
2
=2.8μC=2.8×10
−6
C
Distance between the two spheres =40cm=0.4m
For the mid-point r
1
=r
2
=
2
0.40
=0.2m
Potential at O
V=V
1
+V
2
=
4πε
0
1
[
r
1
q
1
+
r
2
q
2
]=
0.2
9×10
9
(1.8+2.8)×10
−6
=2.1×10
5
V
Explanation:
mark as brainlist
Answer:
Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges.
Magnitude of charge located at A, q1 = 1.5 μC
Magnitude of charge located at B, q2 = 2.5 μC
Distance between the two charges, d = 30 cm = 0.3 m
(a) Let V1 and E1 are the electric potential and electric field respectively at O
V1 = Potential due to charge at A + Potential due to charge at B
Where, ∈0 = Permittivity of free space
E1 = Electric field due to q2 - Electric field due to q1
Therefore, the potential at mid-point is 2.4 × 105 V and the electric field at mid-point is 4 × 105 V m - 1. The field is directed from the larger charge to the smaller charge.
(b) Consider a point Z such that normal distanceOZ = 10 cm = 0.1 m, as shown in the following figure.
V2 and E2 are the electric potential and electric field respectively at Z.
It can be observed from the figure that distance,
V2= Electric potential due to A + Electric Potential due to B
Electric field due to q at Z,
Electric field due to q2 at Z,
The resultant field intensity at Z,
Where, 2θis the angle, ∠AZ B
From the figure, we obtain
Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 2.0 × 105 V and electric field is 6.6 ×105 V m -1
use this formula same