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please help with these two 9. A particle (9 = 4.0 mC, m= 50 g) has...
Part A.) The electric flux through a spherical surface is 4.0\times 10^4~\text{N}\cdot\text{m}^2/\text{C}4.0×10 4 N⋅m 2 /C. What is the net charge enclosed by the surface? Part B.) The electric field 10.0 cm from the surface of a copper ball of radius 5.0 cm is directed toward the ball’s center and has magnitude 9.0*10^2N/C. How much charge is on the surface of the ball? Select the correct answer a.) 2.2*10^{-10} . b.)-1.25*10^{-10} . c.)-1.0* 10^{-9} . d.)-4.5* 10^{-10) ....
Crossed E and B Fields. A particle with initial velocity V = (5.85 * 10^3 m/s) j enters a region of uniform electric and magnetic fields. The magnetic field in the region is B= -(1.350 T) k . Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) +0.640 nC and (b) -0.320 nC. You can ignore the weight of the particle.
A particle with initial velocity T = (6 x 10 m/s) i enters a region of uniform electric and magnetic fields. The magnetic field in the region is B = -(1.5T). In which of the following options is the magnitude and direction of the electric field in the region given correctly if a particle of charge q = 5 nC is to pass through undeflected with a constant velocity? (You can ignore the weight of the particle.) 9(kV/m) 4 (kV/m)...
Please help !!! Calculate the magnitude of the flux of a constant electric field of 5.0 NC in the z -direction through a rectangle with area 4.0 m2 in the xy plane. a. 0 b. 5.0 N-m2/C c. 10.0 N-m2/C d. 20.0 N-m/C 10. 11. A circular ring of charge of radius b has a total charge q uniformly distributed around it. The electric field at the center of the ring is a. 0 12. A solid spherical conductor has...
5. A small particle of velocity vo-8.1 × 103 m/s asit enters a region of uniform magretic field. The particle is observed to travel in the semicircular path with radius R 5.0 cm. Cakculate the (a) magnitude and (b) direction of the magnetic fleld in the region charge q-1,9 x 10-6 C and mass m "3.1x10-12kg has
A negatively charged particle, which has a charge of -50 JC and a mass of 5x10-6 kg, had entered a region of uniform magnetic field with velocity V in the positive x direction before it exited a second region of uniform electric field with velocity-V/2 in the negative x direction as shown in the figure below. if the magnetic field B=1 T and the distance d=10 cm, what was the initial speed V of the particle? v/2 E h x...
A particle with charge of +5.0 X 10^-4 C, and mass of 2.0X10^-9 kg moves at 2.0X10^3 m/s in the +x direction. It enters a uniform magnetic field of 0.20 T that?s points in the +y direction. (a) What will be magnitude and direction of the force on this charged particle due to the magnetic field? Suppose magnetic field exists over a large area, large enough to trap the charged particle and continues to gyrate in the same circle. (b)...
A particle (q= 5.0 nC, m = 3.0 mg) enters a region where the magnetic field has components By = 0, By = 0 and B, = 4.0 T. The initial velocity of the particle as it enters this region is given by Vx - 7.0 m/s, Vy = 6.0 m/s and vz = 8.0 m/s. (a.) What is the magnitude of the magnetic force and the magnitude of the acceleration of the particle (b.) What is the radius and...
A particle (q = 5.0 nC, m = 3.0 mg) enters a region where the magnetic field has components Bx = 0, By = 0 and B. = 4.0 T. The initial velocity of the particle as it enters this region is given by Vx = 7.0 m/s, Vy = 6.0 m/s and V2 = 8.0 m/s. (a.) What is the magnitude of the magnetic force and the magnitude of the acceleration of the particle? (b.) What is the radius...
Help and explain please A particle of a mass 1.2 x 10-11 kg and negative charge - 1.0 uC moves to the right with a speed of 3.0 x 103 m/s. It enters a uniform electric field region where the field is known to be parallel to particle's direction of motion. If the particle is to be brought to rest in the space of 4.0 cm, (a) What potential difference does it have to go through to stop this way?...