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The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.
Answer:
4 m/s
Explanation:
m1 = m2 = m
u1 = 20 m/s, u2 = - 12 m/s
Let the speed of composite body is v after the collision.
Use the conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
m x 20 - m x 12 = (m + m) x v
20 - 12 = 2 v
8 = 2 v
v = 4 m/s
Thus, the speed of teh composite body is 4 m/s.
Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;

Where;
d is the separation or distance between the two parallel plates;

Therefore, the necessary separation between the two parallel plates is 0.104 mm