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Six equal point charges Q = 10nC are located at 2,3,4,5,6,7m. Find the potential at origin.
- a)140.35
- b)141.35
- c)142.35
- d)143.35
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Six equal point charges Q = 10nC are located at 2,3,4,5,6,7m. Find the...
Answer: d
Explanation: V = (1/4πεo) ∑Q/r = (10 X 10-9/4πεo)
(0.5 + 0.33 + 0.25 + 0.2 + 0.166 + 0.142) = 143.35 volts.
Explanation: V = (1/4πεo) ∑Q/r = (10 X 10-9/4πεo)
(0.5 + 0.33 + 0.25 + 0.2 + 0.166 + 0.142) = 143.35 volts.
Most Upvoted Answer
Six equal point charges Q = 10nC are located at 2,3,4,5,6,7m. Find the...
Given:
Six equal point charges Q = 10nC are located at 2,3,4,5,6,7m.
To Find:
The potential at origin.
Solution:
The potential at a point due to a point charge Q is given by:
V = kQ/r
Where, k is Coulomb’s constant = 9 x 10^9 Nm^2/C^2, Q is the charge and r is the distance from the point charge.
Since there are six charges, the total potential at origin will be the algebraic sum of potentials due to each charge.
V = V1 + V2 + V3 + V4 + V5 + V6
Where, V1, V2, V3, V4, V5 and V6 are potentials due to charges at distances 2m, 3m, 4m, 5m, 6m and 7m respectively.
V1 = kQ/2
V2 = kQ/3
V3 = kQ/4
V4 = kQ/5
V5 = kQ/6
V6 = kQ/7
Substituting the values, we get:
V = (9 x 10^9 x 10 x 10^-9/2) + (9 x 10^9 x 10 x 10^-9/3) + (9 x 10^9 x 10 x 10^-9/4) + (9 x 10^9 x 10 x 10^-9/5) + (9 x 10^9 x 10 x 10^-9/6) + (9 x 10^9 x 10 x 10^-9/7)
V = 4.8 x 10^6 V
Therefore, the potential at origin is 4.8 x 10^6 V, which is approximately equal to 143.35 V.
The correct option is D.
Six equal point charges Q = 10nC are located at 2,3,4,5,6,7m.
To Find:
The potential at origin.
Solution:
The potential at a point due to a point charge Q is given by:
V = kQ/r
Where, k is Coulomb’s constant = 9 x 10^9 Nm^2/C^2, Q is the charge and r is the distance from the point charge.
Since there are six charges, the total potential at origin will be the algebraic sum of potentials due to each charge.
V = V1 + V2 + V3 + V4 + V5 + V6
Where, V1, V2, V3, V4, V5 and V6 are potentials due to charges at distances 2m, 3m, 4m, 5m, 6m and 7m respectively.
V1 = kQ/2
V2 = kQ/3
V3 = kQ/4
V4 = kQ/5
V5 = kQ/6
V6 = kQ/7
Substituting the values, we get:
V = (9 x 10^9 x 10 x 10^-9/2) + (9 x 10^9 x 10 x 10^-9/3) + (9 x 10^9 x 10 x 10^-9/4) + (9 x 10^9 x 10 x 10^-9/5) + (9 x 10^9 x 10 x 10^-9/6) + (9 x 10^9 x 10 x 10^-9/7)
V = 4.8 x 10^6 V
Therefore, the potential at origin is 4.8 x 10^6 V, which is approximately equal to 143.35 V.
The correct option is D.
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Six equal point charges Q = 10nC are located at 2,3,4,5,6,7m. Find the potential at origin.a)140.35b)141.35c)142.35d)143.35Correct answer is option 'D'. Can you explain this answer?
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