Question

The electric potential due to a charge at any point is V=(\(3x^2+5\)). The electric field intensity at a point (-2, 1, 0) is:
a)17 V/m
b)-17 V/m
c)+12 V/m
d)-12 V/m

Answer 1

Solve as E=-dV/dx as potential does not change with y and z
replace only for x=-2, then: E=-dV/dx=-6x
E(x=-2)=12 V/m, assuming V is given in volts and x in meters

Answer 2

So, if the relation would be in terms of y; x and z would have nothing to do??

Answer 3

Exactly. E=-((dV/dx)i+(dV/dy)j+(dV/dz)k) will take the value 0 for the components that do not appear in the expression of potential

Answer 4

what if V=(3x^2 +5y^2)?? How would we do it?
would it be E=-(6x+10y)=-(-12+10)=2V/m ??

Answer 5

Nope, you have to take partial derivatives for each component. Say, for component x you calculate the derivative assuming y=constant. Therefore, it would be:
E=-6xi-10yj

Answer 6

Then the Electric field would be:12i-10j but you cannot add the components (remember E is a vector)

Answer 7

Yes, but remember that the vector has the origin in the point and that y-component is negative. The module would be SQROOT(36+100)

Answer 8

It should be \(\sqrt{12^2 +10^2}\) V/m.. right?

Answer 9

Yes, I have mistaken the coeefficients instead the true components, it is SQROOT (244)

Answer 10

thanks buddy!! thanks a lot! :)

Answer 11

welcome!