To determine the electric field from the given potential function, we need to use the relationship between electric field
E and electric potential V. The electric field is the negative gradient of the potential:
E=−∇VGiven the potential function
V(x,y,z)=6xy−y+2yz, we need to compute the partial derivatives of V with respect to
x,y, and z .
First, calculate the partial derivative with respect to x :
‌=‌(6‌xy−y+2‌yz)=6y Next, calculate the partial derivative with respect to
y :
‌=‌(6xy−y+2yz)=6x−1+2z Finally, calculate the partial derivative with respect to z
‌=‌(6xy−y+2yz)=2y Now, the components of the electric field are:
‌Ex=−‌=−6y‌Ey=−‌=−(6x−1+2z)‌Ez=−‌=−2y We are asked to find the electric field at the point
(1,0,1).
Substitute
x=1,y=0, and
z=1 into the components of the electric field:
‌Ex(1,0,1)=−6(0)=0‌Ey(1,0,1)=−(6(1)−1+2(1))=−(6−1+2)=−(7)=−7
‌Ez(1,0,1)=−2(0)=0 Therefore, the electric field at the point
(1,0,1) is:
E=0i−7j+0k=−7jHence, the correct option is:
Option A: -7 j