Electric potential at a point due to two charges

Click For Summary
The discussion focuses on calculating the electric potential at a point due to two charges using the formula V=KQ/r. The user correctly calculated the potentials from each charge, obtaining 36,000V and 18,000V, leading to a total of 54,000V. A clarification was made that the question pertains to electric potential, not electric field, which is a vector quantity. The user acknowledged the oversight in reading the question and confirmed the final answer. This highlights the importance of understanding the distinction between electric potential and electric field in physics problems.
Schaus
Messages
118
Reaction score
5

Homework Statement


What is the electric potential at P due to charges Q1 and Q2?
https://lh3.googleusercontent.com/tUDlHBQ52PYnzufzP-Kow958ph1wtMx-Ar2qhUvAIATr-otVU6ESS6vHnPpv8MNO2NQnII4SQjJ041O9YwGa0LjXHh_rJISBvX7Aoa1PPX7G-YSIQBH6mhyJh0zX2k40M6MceWdCGc7XEMUIT4cnvUqd2s2I6PFbFFBq9NAFftkITBAMNcpevIkdiR4BjYP03Ap0lDGHt3QsM1BNSY4eFannXQbq5WKipGUMk24d4yyewzh2lHph_kOBmCyF3oyTonsfD0hLdpTQeRL6_l4VdlAyFUCHmwcID8vVRNXImlVwsFdQRPO2ZjXog3Fj7HATJbqrhRPNI15vUO99Nn82a3oT26ogHn8jPK5CICvLk8eZZdYfPJuFUOJLCL4xw8_ZVLMZS8xdYw6y4KOg87pN642r8D3sewHjeJejKvtWFkOf6Th5M_kLOXMN2mkMasYc2PoF1rEYNnbyC13KaRZ7uFc9tnK6cDLdLBvAfbwZdjhQ50XZzoc9oK0mWvJIRkK9-rTOHDHlyAEYP8JHfyeSQp7Xg4LX5RA6m9ZFLFAJVDGFGTo1VafxTUlEzs9f-uCOJz6nkO2gQY0z1pcPk3K4NNgps-KCU2YKFkmbgiNmihdEbrtghvgN=w684-h241-no
upload_2016-12-23_14-51-15.png

Homework Equations


V=KQ1/r

The Attempt at a Solution


I used the above formula to...
(9.0 x 109)(4.0 x 10-6)/1 = 36000V
(9.0 x 109)(1.0 x 10-6)/0.5 = 18000V
Now the answer is 54000V and I'm just wondering if that's the answer because you take 36000J + 18000V = 54000V?
 
Last edited by a moderator:
Physics news on Phys.org
Check the equation you're using for electric potential. The question asks for the potential, not the electric field (which would be a vector quantity).
 
  • Like
Likes Schaus
Wow I don't know how I missed the fact it said potential. I may need to take a break. Thanks for your help! I got the answer.
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
View full post »

Similar threads

At what point is Electric Potential zero
  • · Replies 2 ·
Replies
2
Views
2K
Electric potential due to collection of charged particles
  • · Replies 3 ·
Replies
3
Views
2K
Electric Potential Due to point charges
  • · Replies 1 ·
Replies
1
Views
5K
Electric Potential due to two charges
  • · Replies 7 ·
Replies
7
Views
5K
Calculate Net Electric Field at Point P | Two Charges 42cm Apart
  • · Replies 4 ·
Replies
4
Views
2K
Electric field with three charges and a point
  • · Replies 1 ·
Replies
1
Views
3K
Coulomb's Law Grade 12 Question -- Net Electric Field affecting a Charge
  • · Replies 18 ·
Replies
18
Views
4K
Electric Field due to Point Charges
  • · Replies 2 ·
Replies
2
Views
2K
Electric Field due to multiple point charges
  • · Replies 2 ·
Replies
2
Views
4K
What Is the Electric Potential Due to a Line Charge Along Specific Points?
  • · Replies 8 ·
Replies
8
Views
2K