The electric potential V is given as a function of distance ×( metre ) by V=(5x2+10x−4) volt.Question: Value of electric field at x=1 m
Question
The electric potential V is given as a function of distance × (metre) by
V = (5x^2 + 10x - 4) volt.
Question: Value of electric field at x = 1 m
Solution
The electric field E is related to the electric potential V by the relation:
E = -dV/dx
where dV/dx is the derivative of the potential with respect to distance.
Given V = 5x^2 + 10x - 4, we first need to find the derivative of V with respect to x.
dV/dx = 10x + 10
Now, we can find the electric field at x = 1 m by substituting x = 1 into the equation for E:
E = -dV/dx = -(10*1 + 10) = -20 N/C
So, the electric field at x = 1 m is -20 N/C.
Similar Questions
If the electric potential along the x-axis is V = 500 x2𝑥2 V, where x is in meters, what is Ex𝐸𝑥 at x = 0.500 m? 250. V/m -250. V/m -500. V/m 500. V/m
Potential in the x-y plane is given as volts. The electric field at the point (1,-2) will be3J V/m-5 J V/m5 J V/m-3 J V/m
Determine the electric potential produced by a point charge of 6.80x10-7 C at a distance of 2.60 cm.
The variation of electric potential with distance d from a fixed point is as shown in the figure. The electric intensity at d = 5m is
A 4.40 μC charge is subject to a 6.00 mN force due to an electric field. What is the magnitude of the electric field at the location of the charge?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.