The electric potential existing in space is `V(x,y,z)= A (xy+ yz+zx)`.(a) Write the dimensional A. (b) find the expression for the electric field.( c ) If A is 10 SI units, find the magnitude of the electric field at (1m, 1m, 1m).

Explanation

The formula E = `(-dv)/(dr)` in extended form is written as follows :
`vec(E ) = - [ hat(i) (del V)/(del x) + hat(j) (del V)/(del y) + hat(k) (del V)/(del z)]` where, `(del V)/(del x), (del V)/(del y ) "and" (del V)/(del z)` are partial derivatives of potential V with respect to x,y and z respectively . for partial derivative you should note that is say you are calculating `(del V)/(del x)` , then other variables y and z will be treated as constant .
Now, if we write `vec(E) = hat(i)E_(x) + hat(j)E_(y) + hat(k)E_(z)`, then `E_(x) = (-del V)/(del x) , E_(y) = (-del V)/(del y) ad E_(z) = (-del V)/(del z) .` in the question V = B( xy + yz + zx )
so, `E_(x) = (-del V)/(del x) = - B(y + z) ` becasue variables y and z are treated constant.
Now, put the values of B, y and z. Given B = 10 and (x, y , z) are (1,1,1) respecitvely.
So, `E_(x) = (- del V)/(del x ) = - B(y + z)` = -10(1+1) = - 20 N/C
Similarly, `E_(y) = (- del V)/(del y ) = - B(x + z)` [ now treating x and z constant ]
therefore `E_(y) = - 10(1 + 1) = - 20 ` N/S
and `E_(z) = (-del V)/(del z) = ` - B (x + y ) = - 20 N/C
therefore, `vec(E ) = E_(x)hat(i) + E_(y)hat(j) + E_(z)hat(k)`
or `vec(E ) = (-20 hat(i) - 20 hat(j) - 20 hat(k))` N/C
Now, if we want the magnitude of `vec(E ),` then it's equal to
`sqrt(E_(x)^(2) + E_(y)^(2) + E_(z)^(2)) = sqrt((-20)^(2) + (-20)^(2) + (-20)^(2)) = 20sqrt(3)`