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Wize University Physics Textbook (Master) > Electric Potential and Potential Energy

Relationship Between Electric Field and Potential (with calculus)

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Relationship Between Electric field and Potential (with calculus)


As you might expect, anytime rates of change are involved, we can use calculus to simplify some problems and approach more challenging ones.
  • In one dimension, the relationship between electric field and potential can be written with a derivative as follows:
E=−dVdr\boxed{E=-\frac{dV}{dr}}
  • In more than one dimension, we have to consider the rate of change in each direction separately. For this, we use partial derivatives:
Ex=−∂V∂x,   Ey=−∂V∂y,   Ez=−∂V∂z\boxed{E_x=-\frac{\partial V}{\partial x},~~~E_y=-\frac{\partial V}{\partial y},~~~E_z=-\frac{\partial V}{\partial z}}
  • Because we are taking a scalar quantity (potential) and turning it into a vector quantity (field), we can also introduce the gradient, which is a more concise way of writing the above three equations:
E⃗=−∇⃗V\boxed{\vec E=-\vec \nabla V}
  • We can also use definite integrals to determine the electric potential difference between two points if we know the electric field:
Vb−Va=−∫abE⃗⋅dr⃗\boxed{V_b-V_a=-\int_a^b \vec E \cdot d\vec r}

Wize Tip
Gauss's Law is a powerful tool to find the electric field strength for several symmetric charge distributions. There is no analogous tool for the electric potential. However, using these relationships, it is possible to easily obtain the electric potential for a symmetric charge distribution after we found the electric field with Gauss's Law.

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Practice: Determine Electric Field from a Potential


The electric potential in a region is given by the equation V(x,y,z)=2xy+3xz−z2V(x,y,z)=2xy+3xz-z^2. Find the magnitude of electric field at the point (1,2,−1)(1,2,-1).

Electric field is related to electric potential through the following equation:
E⃗=−∇⃗V=−(∂V∂xi^+∂V∂yj^+∂V∂zk^)\vec{E}=-\vec{\nabla}V=-\bigg(\cfrac{\partial V}{\partial x}\hat{i}+\cfrac{\partial V}{\partial y}\hat{j}+\cfrac{\partial V}{\partial z}\hat{k}\bigg)

Taking partial derivatives to find the component of electric field:
∂V∂x=2y+3z         ∂V∂y=2x         ∂V∂z=3x−2z\cfrac{\partial V}{\partial x}=2y+3z\ \ \ \ \ \ \ \ \ \cfrac{\partial V}{\partial y}=2x\ \ \ \ \ \ \ \ \ \cfrac{\partial V}{\partial z}=3x-2z
At the (x,y,z)=(1,2,-1), the magnitude of each component will be:
∂V∂x=2(2)+3(−1)=1         ∂V∂y=2x=2(1)=2         ∂V∂z=3(1)−2(−1)=5\frac{\partial V}{\partial x}=2(2)+3(-1)=1\ \ \ \ \ \ \ \ \ \frac{\partial V}{\partial y}=2x=2(1)=2\ \ \ \ \ \ \ \ \ \frac{\partial V}{\partial z}=3(1)-2(-1)=5
The electric field at the point is:
E⃗=−(1i^+2j^+5k^)\vec{E}=-(1\hat{i} +2\hat{j}+5\hat{k})
The electric field magnitude is found as follows:
∣E⃗∣=(−1)2+(−2)2+(−5)2=30 N/C|\vec{E}|=\sqrt{(-1)^2+(-2)^2+(-5)^2}=\sqrt{30}~N/C

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Practice: Electric Potential and Potential Energy of a Conducting Sphere


Consider a metallic, solid sphere of radius RR.

a) Find the electric potential V of the sphere (inside and outside).
b) Find the electric potential energy U.

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Practice: Electric Potential and Potential Energy of an Insulating Sphere


Consider a solid, insulating sphere of radius RR and uniform charge density ρ\rho.

a) Find the electric potential V of the sphere (both inside and outside).
b) Plot your results as a function of rr in addition to the plot of the conducting case. To make plotting easier, you can set all constants equal to 1 (Q, 4π4\pi, ϵo\epsilon_o, R).
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Practice: Electron Moving Away from Cylindrical Insulator


An infinitely long, solid, cylindrical insulator has a uniform linear charge density of 5.0 Cm5.0~ \frac{C}{m}and a radius of2.0 cm2.0 ~cm. An electron with an initial velocity of 4.0×1010ms4.0\times10^{10} \frac{m}{s} is initially 10.0 cm10.0~cm away from the axis of the cylinder and is moving away from the cylinder on a path that is perpendicular to the axis of the cylinder.

At what distance away from the axis of the cylinder will the speed of the electron be zero?

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Practice: Electric Potential of a Spherical Conducting Shell


Consider a spherical metallic shell of inner radius 3.0 cm, outer radius 5.0 cm, and net charge -10 C, with a point charge of +2.0 C placed inside the shell at the center of the cavity.


a) One probe of a voltmeter is placed at the outer surface of the metallic shell, and the other probe is placed a distance 10 cm away from the center of the shell. What is the reading of the voltmeter?
b) What is the potential of the inner surface of the shell (assume the voltage at infinity is set to be zero)?

Hint: Remember that the electric field of this geometry (inside or outside of the shell) follows the same behaviour as a single point charge.