The electric potential V in the space between two flat parallel plates 1 and 2 is given (in volts) by V = 2100x², where x (in meters) is the perpendicular distance from plate 1. At x = 1.3 cm, (a) what is the magnitude of the electric field and (b) is the field directed toward or away from plate 1? (a) N/C (b)

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**Electric Potential and Electric Field Between Parallel Plates**

In this exercise, we will explore the electric potential V in the space between two flat parallel plates labeled Plate 1 and Plate 2. The electric potential V (in volts) is given by the equation:

\[ V = 2100x^2 \]

where \( x \) (in meters) is the perpendicular distance from Plate 1. We are asked to determine:

1. The magnitude of the electric field at a distance \( x = 1.3 \) cm.
2. The direction of the electric field, specifically if it is directed toward or away from Plate 1.

### Steps to Solve:

**(a) Magnitude of the Electric Field**

To find the electric field (E), we need to take the derivative of the electric potential (V) with respect to x. The relationship between the electric field and electric potential is given by:

\[ E = -\frac{dV}{dx} \]

Given the electric potential function:

\[ V = 2100x^2 \]

The derivative of \( V \) with respect to \( x \) is:

\[ \frac{dV}{dx} = 4200x \]

Substituting \( x = 1.3 \) cm (which is \( x = 0.013 \) meters):

\[ E = -4200 \times 0.013 = -54.6 \text{ N/C} \]

The magnitude of the electric field at \( x = 1.3 \) cm is:

\[ |E| = 54.6 \text{ N/C} \]

**(b) Direction of the Electric Field**

From the equation \( E = -4200x \), it is evident that if \( x \) is positive, \( E \) will be negative, indicating that the electric field is directed towards lower potential, which means it is directed toward Plate 1.

Therefore, at \( x = 1.3 \) cm, the field is directed toward Plate 1.

To summarize:
- The magnitude of the electric field at \( x = 1.3 \) cm is 54.6 N/C.
- The electric field is directed toward Plate 1.

This concludes the problem, providing us with both the magnitude and direction of the electric field between the parallel plates.
Transcribed Image Text:**Electric Potential and Electric Field Between Parallel Plates** In this exercise, we will explore the electric potential V in the space between two flat parallel plates labeled Plate 1 and Plate 2. The electric potential V (in volts) is given by the equation: \[ V = 2100x^2 \] where \( x \) (in meters) is the perpendicular distance from Plate 1. We are asked to determine: 1. The magnitude of the electric field at a distance \( x = 1.3 \) cm. 2. The direction of the electric field, specifically if it is directed toward or away from Plate 1. ### Steps to Solve: **(a) Magnitude of the Electric Field** To find the electric field (E), we need to take the derivative of the electric potential (V) with respect to x. The relationship between the electric field and electric potential is given by: \[ E = -\frac{dV}{dx} \] Given the electric potential function: \[ V = 2100x^2 \] The derivative of \( V \) with respect to \( x \) is: \[ \frac{dV}{dx} = 4200x \] Substituting \( x = 1.3 \) cm (which is \( x = 0.013 \) meters): \[ E = -4200 \times 0.013 = -54.6 \text{ N/C} \] The magnitude of the electric field at \( x = 1.3 \) cm is: \[ |E| = 54.6 \text{ N/C} \] **(b) Direction of the Electric Field** From the equation \( E = -4200x \), it is evident that if \( x \) is positive, \( E \) will be negative, indicating that the electric field is directed towards lower potential, which means it is directed toward Plate 1. Therefore, at \( x = 1.3 \) cm, the field is directed toward Plate 1. To summarize: - The magnitude of the electric field at \( x = 1.3 \) cm is 54.6 N/C. - The electric field is directed toward Plate 1. This concludes the problem, providing us with both the magnitude and direction of the electric field between the parallel plates.
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