Answer:
A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in the following figure. The particle, confined to move along the x axis, is moved a small distance x along the axis ( where x << a) and released. Show that the particle oscillates in simple harmonic motion with a frequency given by,
I believe the correct answer from the choices listed above is option B. In a food chain. it is the producers trophic level that contains <span>the greatest amount of energy since it is the very start of the chain. Most of the energy are being stored and passed to the next organism. Hope this answers the question.</span>
Answer:
axial V = 0
equatorial V = k q 2a / (x² -a²), V = k q 2x / (a² -x²)
Explanation:
A dipole is a system formed by two charges of equal magnitude, but different sign, separated by a distance 2a; let's look for the electrical potential in an axial line
V = k (q / √(a² + y²) - q /√ (a² + y²))
V = 0
the potential on the equator
we place the positive charge to the left and perform the calculation for a point outside the dipole
V = k (q / (x-a) - q / (x + a))
V = k q 2a / (x² -a²)
we perform the calculation for a point between the dipo charges
V = k (q / (a-x) - q / (a + x))
V = k q 2x / (a² -x²)
The answer is false hope this helps
