Poll Results
No votes. Be the first one to vote.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
Answer: c
Explanation: V = -∫ E.dl = -∫ (40xy dx + 20×2 dy + 2 dz) , from q to p.
On integrating, we get 106 volts
To find the potential difference between two points ( p(1, -1, 0) ) and ( q(2, 1, 3) ) in the electric field ( mathbf{E} = 40xy mathbf{i} + 20x^2 mathbf{j} + 2 mathbf{k} ), we can use the relationship between the electric field and electric potential ( V ):
[
V = – int mathbf{E} cdot dmathbf{r}
]
The potential difference between two points ( V(p) – V(q) ) can be calculated using a path integral from point ( p ) to point ( q ).
1. Determine the path of integration: A simple path can be taken along the coordinate axes. For example, move from ( p(1, -1, 0) ) to ( q(2, -1, 0) ), then to ( q(2, 1, 0) ), and finally to ( q(2, 1, 3) ).
2. Calculate the integrals along each segment:
– Segment 1: From ( p(1, -1, 0) ) to ( (2, -1, 0) ):
[
dmathbf{r} = dx mathbf{i} + 0 mathbf{