A 12 cm×12 cm rectangle lies in the first quadrant of the xy-plane with one corner at the origin. Unit vector nˆ points in the +𝒵 -direction. What is the electric flux through the rectangle if the electric field is E (→ above E) = (2000 m¯¹) x kˆ N/C? Hint: Divide the rectangle into narrow strips of width .

Verified step by step guidance

1. Identify the direction of the electric field and the area vector of the rectangle. The electric field is given as \( \vec{E} = 2000x \hat{k} \) N/C, which means it varies with x and points in the positive z-direction. The area vector \( \vec{A} \) of the rectangle also points in the positive z-direction (\( \hat{k} \)) since the normal vector \( \hat{n} \) is given to point in the +z-direction.
2. Calculate the area of the rectangle. The area \( A \) of a rectangle is given by the product of its length and width. For a square with side 12 cm, the area is \( 12 \text{ cm} \times 12 \text{ cm} = 144 \text{ cm}^2 \). Convert this area into square meters (m²) by using the conversion factor (1 m = 100 cm).
3. Set up the integral to calculate the electric flux. The electric flux \( \Phi \) through the rectangle is given by the integral of the dot product of the electric field and the differential area vector over the surface, \( \Phi = \int \vec{E} \cdot d\vec{A} \). Since \( \vec{E} \) and \( d\vec{A} \) are both in the z-direction, the dot product simplifies to the product of their magnitudes.
4. Express the differential area vector in terms of its dimensions. Divide the rectangle into narrow strips parallel to the y-axis, each with a width dx. The differential area vector for each strip is \( d\vec{A} = dx \times 12 \text{ cm} \times \hat{k} \).
5. Integrate the expression for electric flux from x = 0 to x = 12 cm. Substitute \( \vec{E} = 2000x \hat{k} \) and \( d\vec{A} = 12 \text{ cm} \times dx \times \hat{k} \) into the integral and perform the integration over the limits from 0 to 12 cm (converted to meters). This will give you the total electric flux through the rectangle.