A charge emanates an electric field of 97 N/C. If this field is placed on a circular surface with a radius of 53 um & is angled at 69 degrees from the surface, what is the electric flux?
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A charge emanates an electric field of 97 N/C. If this field is placed on a circular surface with a radius of 53 um & is angled at 69 degrees from the surface, what is the electric flux?
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- A constant electric field of magnitude Eo points in the direction of the positive z-axis. What is the net electric flux through a cube, given the area (at z=0 and z>0) normal to the electric fields is A? Eo O 0 O 2*Eo*A +Eo*A -Eo*AA uniform electric field of magnitude E=435 N/C makes an angle of θ= 65.0 ° with a plane surface of area A=3.50m² as in the figure below. What is the electric flux through the this surface?An electric field of 500 V/m makes an angle of 30.00 with the surface vector. It has a magnitude of 0.500 m2. Find the electric flux that passes through the surface.
- A box with a cross-section of 1m x 1m x 1m is surrounding a plan of charge. The uniform electric field is equal to 8 N/C. The total electric flux through this box isA uniform electric field of magnitude 5.7 x 104 N/C is at an angle of 10° to a square sheet with sides 5.5 m long. What is the electric flux through the sheet? Hint Electric flux is E N.m²/C.A uniform electric field of magnitude 3.6×104N/C is at an angle of 80° to a square sheet with sides 8.5 m long. What is the electric flux through the sheet?
- An electric point charge of Q = 21.3 nC is placed at the center of a sphere with a radius of r = 46.1 cm. The sphere in this question is only a mathėmatical surface, it is not made out of any physical material. What 'is the electric flux through the surface of this sphere? Submit Answer Tries 0/99 This same point charge is now moved out from the center of the sphere by a distance of 18.9 cm. What is the electric flux through the surface of the sphere now? Submit Answer Tries 0/99An electric field 7 N/C is passing through a region of area 2.1 m² in the xy-plane. If the electric field makes an angle of 56° with respect to the area vector perpendicular to the region, calculate the electric flux magnitude (in N-m²/C) through the region.An electric flux of 18.35 Nm2/C passes through a flat surface that is perpendicular to a constant electric field of strength 11.5 N/C. What is the area of the surface?
- A thundercloud produces a vertical electric field of magnitude 2800N/C at ground level. You hold 22.0cm x 28.0cm sheet of paper horizontally below the cloud. (a) What is the electric flux through the sheet?(b) What would the flux be if you tilt the sheet of paper by 30°? (c) What would the flux be if you hold the sheet of paper vertically?For a given surface, the electric flux, PE, is proportional to the number of field lines through the surface. For a uniform electric field, the maximum electric flux is equal to the product of electric field at the surface and the surface area (i.e., EA). The electric flux is defined to be positive when the electric field E has a component in the same direction as the area vector A and is negative when the electric field has a component in the direction opposite to the area vector. C. Sketch vectors A and Ē such that the electric flux is: Positive Negative ZeroCalculate the electric field in N/C at point P, a distance (4.35x10^1) cm along the central axis of a disk of charge with radius (9.157x10^0) cm, and charge density +(8.0860x10^0) µC/m2. You do not need to enter a unit vector in your answer, but must put a negative sign in, if the electric field is pointing along the negative z-axis. R