EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 8220100663987
Author: Jewett
Publisher: Cengage Learning US
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Chapter 24, Problem 24.49AP
Find the electric flux through the plane surface shown in Figure P23.37 if θ = 60.0°, E = 350 N/C, and d = 5.00 cm. The electric field is uniform over the entire area of the surface.
Figure P23.37
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Consider the uniform electric field E = (3.5 j + 3.5 k) × 103 N/C.
a) Calculate the electric flux through a circular area of radius 1.75 m that lies in the yz-plane. Give your answer in N·m2/C.
b) Repeat the electric flux calculation for the circular area for the case when its area vector is directed at 45° above the xy-plane. Give your answer in N·m2/C.
A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux through the
loop is a maximum. At this position, the electric flux is 7.50 x 105 N-m²/C. Determine the magnitude of the electric field.
O 2.44 × 106 N/C
O 4.24 × 106 N/C
O 1.07 x 106 N/C
O 8.88 x 105 N/C
O.6.00 x 106 N/C
eTextbook and Media
Figure P15.49 shows a closed
cylinder with cross-sectional
area A = 2.00 m². The con-
stant electric field É has mag-
nitude 3.50 x 10% N/C and
is directed vertically upward,
perpendicular to the cylinder's
top and bottom surfaces so that no
field lines pass through the curved
surface. Calculate the electric flux
through the cylinder's (a) top and
(b) bottom surfaces. (c) Determine
the amount of charge inside the
cylinder.
Figure P15.49
Chapter 24 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
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- A uniform electric field of magnitude E = 38.5 N/C makes and angle of θ = 67.3° relative to a line normal to a surface with area A = 3.01 m2. What is the electric flux (Φ) that passes through this surface? Your answer should be in N·m2/Carrow_forwarda circular surface with a radius of 0.061 m is exposed to a uniform electric field of magnitude 1.88E4 N/C. The electric flux through the surface is 77 Nm^2/C. What is the angle between the diraction of the electric field and the normal to the surfacearrow_forwardConsider the uniform electric field E = (3.5 jˆ + 3.5 kˆ) × 103 N/C. A) Calculate the electric flux through a circular area of radius 1.75 m that lies in the yz-plane. Give your answer in N·m2/C. B) Repeat the electric flux calculation for the circular area for the case when its area vector is directed at 45° above the x-axis (pointing in the direction of iˆ+kˆ). Give your answer in N·m2/C.arrow_forward
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- A cylindrical metal can have a height of 27 cm and a radius of 11 cm. The electric field is directed outward along the entire surface of the can (including the top and bottom), with a uniform magnitude of 4.0x10^5 N/C. a.) What is the surface area of the cylinder?arrow_forwardThere is a rectangle of length 2.70 cm and width 1.50 cm in the xy-plane. If the electric field can be expressed by (50.0i^i^ +160. k^k^) N/C, what is the flux through the rectangle? 0.0648 N m2/C 0 N m2/C 0.0203 N m2/C 0.0679 N m2/Carrow_forwardAn insulating solid sphere of radius R = 6.0cm has a total positive charge Q uniformly distributed throughout its volume. The electric flux through a spherical Gaussian surface of radius r = 3.0cm is 2.26x105N.m2/C. How much charge (in units of μC) is enclosed by the Gaussian surface of radius r =3.0cm? What is the magnitude ( in units of 106N/C) of the E-field at the Gaussian surface of part (1)? What is the magnitude (in units of 106N/C) of the E-field at surface of the sphere?arrow_forward
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