Answer:
Explanation:
Radius of ring r = .05 m
Electric field due to a uniformly charged ring
E = k Q x / ( x² + r² )³/² ; Q is charge on the ring , x is distance of point from the center of the ring and r is radius of the ring
electric field due to first ring at middle point or at x = 7.5 cm
E = 9 x 10⁹ x 33 x 10⁻⁹ x 7.5 x 10⁻² / ( 7.5² + 5² )³/² x 10⁻³
= 9 x 10⁹ x 33 x 10⁻⁹ x 7.5 / ( 7.5² + 5² )³/² x 10⁻¹
= 9 x 10⁹ x 33 x 10⁻⁹ x 7.5 / 73.23
= 30.41 N/C
The same field will be created by the other ring at the middle point because charge on the ring is same in magnitude . Due to negative charge on the second ring , field due to both the rings will align in the same direction.
Total field = 2 x 30.41
= 60.82 N/C
Answer:
Four charges of equal magnitude sitting at the vertices of a square
Explanation:
We can arrive at such a situation by thinking of a simple example first, a configuration of two charges. The force acting on the middle point of a straight line joining the two points(charges) will be zero. That is, the net Electric field will be zero as they cancel out being equal in magnitude and opposite in direction.
Now, we can extend this idea to a square having charge q at each vertex. If we put 'p' at the geometric center, we can see that the Electric fields along the diagonals cancel out due to the charges at the diagonally opposite vertices(refer to the figure attached). Actually, the only requirement is that the diagonally opposite charges are equal.
We can further take this to 3 dimensions. Consider a cube having charges of equal magnitude at each vertex. In this case, the point 'p' will yet again be the geometric center as the Electric field due to the diagonally opposite charges will cancel out.
Answer:
False
Explanation:
ac = v^2/r
acceleration is not dependent on the mass of the orbiting object.
Answer:
V=Bh
Explanation:
B h is used for rectangular solids and cylinders