In a certain region of space the electric potential is given by V=+A x^2 y-B x y^2, where A=5.00

step 1 So the electric fuel can be defined by the following equation. So E is equal to the derivative o

In a certain region of space the electric potential is given...

Key Concepts

Electric Potential

Electric potential is a scalar quantity that represents the work done per unit charge in bringing a test charge from infinity to a given point in an electric field. It provides a convenient way to describe the energy landscape of a charge distribution, and changes in the potential relate directly to the forces that charged particles experience.

Electric Field

The electric field is a vector quantity that represents the force per unit charge exerted on a small positive test charge placed in the field. It is directly related to the spatial rate of change of the electric potential and indicates the direction in which a positive charge would accelerate.

Gradient Operator

The gradient operator is a vector differential operator that, when applied to a scalar function, yields the vector field pointing in the direction of the greatest rate of increase of the function. In the context of electric potential, the gradient allows us to find how rapidly and in which direction the potential changes in space.

Partial Derivatives

Partial derivatives involve differentiating a multivariable function with respect to one variable while keeping other variables constant. They are used to compute each component of the gradient of the electric potential, which in turn is used to determine the corresponding components of the electric field.

Vector Magnitude and Direction

The magnitude of a vector quantifies its overall size regardless of its direction, while its direction indicates the line along which it acts. In electric field calculations, after computing the field components, the magnitude is obtained by taking the square root of the sum of the squares of its components, and the direction is often expressed as a normalized unit vector.

Transcript

00:01 So the electric fuel can be defined by the following equation.

00:05 So e is equal to the derivative of the potential over the derivative of the position, r.

00:15 So now if you substitute the potential that we're given, which is v is equal positive ax squared, y minus bx squared, b, y squared.

00:27 So if you substitute that, let me, hold on, let me just get that down on.

00:32 So p is equal to a x squared y minus b x squared so now we see substitute that and bring it into the potential into this equation.

00:51 So let me write that in back.

00:54 So d d r that's a really long line.

01:02 Get rid of that and now go back to pen.

01:07 So now a x minus b x y squared so now if you compute this uh so now we need to compute it for each of the components so if we compute it for e x and the so now let's solve for the position that x so now it's de x so with respect to x and now let me just uh bring this straight down here looks all right uh let me take this here let me delete that and uh let me just make it easier so if we can do this quicker so if you move this down here and uh differentiate it with respect to x we're going to get that that is equal to two times a x y minus and since there's x with power 1 it just goes the course the so it's b y square.

02:29 And now we know the constants which are just given to us in the equation in the question.

02:34 So if you substitute that, like i don't think i need to write it, but if you plug it straight into our calculator, substitute the values that we get from the question, we're going to get that x is equal to negative 6 .72 volts per meter.

02:54 So now that's the potential at x.

02:57 So i mean to calculate the potential at y.

03:00 So if you do the same thing, so the differentiated by this time with this respect to y, since we're calculating the percential at y, if you bring this here, bring this down here.

03:15 So now if you differentiate it with respect to y, we're going to get that equal to since there's y to the power one here, it just becomes the coefficients.

03:23 So a x squared.

03:25 And since there's y to the power two here, it just we bring the two the other side and then subtract the power.

03:31 So it's minus 2bx y.

03:38 And now if we subsidize the values that we get from the questions right into this, say into this formula, we can get that it's equal to negative 7 .2 volts per meter squared.

03:53 And always remember the unit.

03:56 And finally, the potential at c.

04:02 I actually don't think i need to write it out...

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