Question

The potential of the electric field produced by point charge at any point (x, y, and z) is given by \[V = 3{x^2} + 5\] where x, y is in meters and V is in volts. The intensity of the electric field at (−2, 1, 0) is:
A) $ + 17V{m^{( - 1)}}$
B) $ - 17V{m^{( - 1)}}$
C) $ + 12V{m^{( - 1)}}$
D) $ - 12V{m^{( - 1)}}$

Answer 1

Hint: The amount of energy needed to move from one point to another is known as the electric potential field. With the help of the intensity of this field we can solve the given problem.

Formula used:
Use the formula,
\[E = \dfrac{{ - dv}}{{dx}}\]
Where E is the electric field.

Complete step-by-step answer:
The potential of an electric field produced by a point charge is given by \[V = 3{x^2} + 5\]
We can use the intensity of the electric field. The intensity of the electric field is given by, \[E = \dfrac{{ - dv}}{{dx}}\]
We can substitute the given values in the formula.
 \[E = \dfrac{{ - d(3{x^2} + 5)}}{{dx}}\]
By using the differentiation, we can simplify the equation.
\[E = - 6x\]
Now let us substitute the intensity value in the x-value
\[E = - 6( - 2)\]
\[E = + 12V{m^{ - 1}}\]
Therefore, from the question we can find out that, \[E = + 12V{m^{ - 1}}\]

Hence the correct option is (c).

Note: An electric field created due to a charge C at any point in space can be defined as the force faced by a unit positive charge placed on that point. The charge C is called a source charge and is responsible for creating the electric field, while the charge c is called the test charge and is responsible for testing the effects of a source charge.
An electric field line is generally a curve that is drawn so that tangent to its every point is that point’s direction of the net field. The arrows on the curve are necessary and indicate the direction of the field from both possible directions. A field line is similar to a space curve and is three dimensional. The pattern of lines thus formed, are often referred to as electric field lines and they point towards the direction a positive test charge would accelerate after being placed on the line. In this case, the lines will point in the opposite directions of the positively charged source charges and point towards the direction of the negatively charged source charges.
Remember there is a negative sign in the formula \[E = \dfrac{{ - dv}}{{dx}}\]. The negative sign indicates that when E increases velocity tends to decrease.