Answer: higher and lower
Explanation:
charge in an electric field will experience a force in the direction of decreasing potential energy. Since the electric potential energy of a negative charge is equal to the charge times the electric potential the direction of decreasing electric potential energy is the direction of increasing electric potential.
Answer:
Explanation:
Given that,
Frequency of radio signal is
f = 800kHz = 800,000 Hz.
Distance from transmitter
d = 8.5km = 8500m
Electric field amplitude
E = 0.9 V/m
The average energy density can be calculated using
U_E = ½•ϵo•E²
Where ϵo = 8.85 × 10^-12 F/m
Then,
U_E = ½ × 8.85 × 10^-12 × 0.9²
U_E = 3.58 × 10^-12 J/m²
The average electromagnetic energy density is 3.58 × 10^-12 J/m²
Solution :
We know, resistance is given by :


Now, we know mass of wire is given by :

Hence, this is the required solution.
The equator is warmer than the poles because the equator is closer to the sun. In other words, the sun is overhead the equator, which is a result of the Earth's curvature.
Complete Question
Suppose you have three identical metal spheres, A, B, and C. Initially sphere A carries a charge q and the others are uncharged. Sphere A is brought in contact with sphere B, and then the two are separated. Spheres CC and BB are then brought in contact and separated. Finally spheres AA and CC are brought in contact and then separated. What is the final charge on the sphere B, in terms of q?
a. 3/8q
b. 1/4q
c. 3/4q
d. q
e. 5/8q
f. 1/3q
g.1/2q
h. 0
Answer:
The correct option is b
Explanation:
From the question we are told that
The charge carried by A is q C
The charge carried by B is 0 C
The charge carried by C is 0 C
When A and B are brought close and then separated the charge carried by A and B is mathematically evaluated as

When C and B are brought close and then separated the charge carried by C and B is mathematically evaluated as

When C and A are brought close and then separated the charge carried by C and A is mathematically evaluated as

Looking at these calculation we can see that the charge carried by B is
