A cube with sides of area 32 cm2 contains a 35.9 nanoCoulomb charge. Find the flux of the electric field through the surface of
the cube in unis of Nm2/C.
1 answer:
Answer:
The electric flux through the surface is equal to 3.878 x 10³ Nm²/C
The field distance r is equal to half the length of each side of the cube.
From the area the length of each size was calculated and the field distance and charge were used in calculating the magnitude of the electric field vector which was found to be 202 x 10³ N/C
The total flux area available to this electric field is 6x32cm²
Explanation:
The full solution can be found in the attachment below.
Thank you for reading this post and I hope it is helpful to you.
You might be interested in
Answer:
B = 7.6 T direction of + x
Explanation:
For the proton beam to continue in the same direction the electric and magnetic forces must be equal
= 0
= F_{e}
Fm = q E
The electric force is in the direction of the electric field because it is the charge of the positive proton, the electric force goes in the direction of –y, therefore, the magnetic force cancels this force must go in the direction of + y
The magnetic force is
F_{m} = q v x B = q v B sin θ
θ = 90
B = q E / q v
B = E / v
B = 800/105
B = 7.6 T
To find the direction of the magnetic field we use the right hand rule, the thumb goes in the direction of the proton velocity, the fingers extended in the direction of the magnetic field and the palm is the direction of force, for a positive charge.
Thumb goes in the direction of the + z axis
Palm in the direction of +y
Fingers point in the direction of + x