Question 89: Electrostats & Capacitance

Question

A square shaped aluminium coin weighs $$0.75 \mathrm{~g}$$ and its diagonal measures $$14 \mathrm{~mm}$$. It has equal amounts of positive and negative charges. Suppose those equal charges were concentrated in two charges $$(+Q$$ and $$-Q)$$ that are separated by a distance equal to the side of the coin, the dipole moment of the dipole is

Accepted Answer

To determine the electric field from the given potential function, we need to use the relationship between electric field $$\mathbf{E}$$ and electric potential $$\mathrm{V}$$. The electric field is the negative gradient of the potential:
$$\mathbf{E} = -
abla \mathrm{V}$$
Given the potential function $$\mathrm{V}(\mathrm{x}, \mathrm{y}, \mathrm{z})=6 \mathrm{xy}-\mathrm{y}+2 \mathrm{yz}$$, we need to compute the partial derivatives of $$\mathrm{V}$$ with respect to $$\mathrm{x}$$, $$\mathrm{y}$$, and $$\mathrm{z}$$.
First, calculate the partial derivative with respect to $$\mathrm{x}$$:
$$\frac{\partial \mathrm{V}}{\partial \mathrm{x}} = \frac{\partial}{\partial \mathrm{x}}(6 \mathrm{xy} - \mathrm{y} + 2 \mathrm{yz}) = 6 \mathrm{y}$$
Next, calculate the partial derivative with respect to $$\mathrm{y}$$:
$$\frac{\partial \mathrm{V}}{\partial \mathrm{y}} = \frac{\partial}{\partial \mathrm{y}}(6 \mathrm{xy} - \mathrm{y} + 2 \mathrm{yz}) = 6 \mathrm{x} - 1 + 2 \mathrm{z}$$
Finally, calculate the partial derivative with respect to $$\mathrm{z}$$:
$$\frac{\partial \mathrm{V}}{\partial \mathrm{z}} = \frac{\partial}{\partial \mathrm{z}}(6 \mathrm{xy} - \mathrm{y} + 2 \mathrm{yz}) = 2 \mathrm{y}$$
Now, the components of the electric field are:
$$ \mathbf{E}_x = -\frac{\partial \mathrm{V}}{\partial \mathrm{x}} = -6 \mathrm{y} $$
$$ \mathbf{E}_y = -\frac{\partial \mathrm{V}}{\partial \mathrm{y}} = -(6 \mathrm{x} - 1 + 2 \mathrm{z}) $$
$$ \mathbf{E}_z = -\frac{\partial \mathrm{V}}{\partial \mathrm{z}} = -2 \mathrm{y} $$
We are asked to find the electric field at the point $$(1,0,1)$$.
Substitute $$\mathrm{x} = 1$$, $$\mathrm{y} = 0$$, and $$\mathrm{z} = 1$$ into the components of the electric field:
$$ \mathbf{E}_x(1,0,1) = -6(0) = 0 $$
$$ \mathbf{E}_y(1,0,1) = -(6(1) - 1 + 2(1)) = -(6 - 1 + 2) = -(7) = -7 $$
$$ \mathbf{E}_z(1,0,1) = -2(0) = 0 $$
Therefore, the electric field at the point $$(1,0,1)$$ is:
$$\mathbf{E} = 0 \mathbf{i} - 7 \mathbf{j} + 0 \mathbf{k} = -7 \mathbf{j}$$
Hence, the correct option is:
Option A: $$\mathrm{-7 j}$$

Answer

$$34.8 \mathrm{~Cm}$$

Answer

$$3.48 \mathrm{~Cm}$$

Answer

$$3480 \mathrm{~Cm}$$

Answer

$$348 \mathrm{~Cm}$$

Electrostats & Capacitance

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