A (adenine) - T (Thymine)
C (cytosine) - G (Guanine)
Answer: the radial distance between the 500-v equipotential surface and the 1000 v surface will be 8.91*106 times the charge Q.
Explanation: To find the answer, we have to know more about the equipotential surfaces.
<h3>What are equipotential surfaces?</h3>
- An equipotential surface is the locus of all points which have the same potential due to the charge distribution.
- Any surface in an electric field, at every point of which, the direction of electric field is normal to the surface can be regarded as equipotential.
- We have the equation for electric potential as,
, where k is equal to 1/(4π∈₀) = .
- equation for radial distance will be,
<h3>How to solve the problem?</h3>
- For the first surface, we can write the equation of potential as,
- For the second surface, we can write the equation of potential as,
- Thus, the radial distance will be,
Thus, we can conclude that, the radial distance between the equipotential surface of 500V and 1000V will be,8.91*106 times the charge Q.
Learn more about the equipotential surface here:
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All carbon atoms have 6 protons<span> in their nucleus. Most carbon atoms also have </span>6 neutrons<span>, giving them an atomic mass of 12 = </span>6 protons<span> + </span>6 neutrons<span>. C-14 atoms have two extra neutrons, giving them a total of </span>8 neutrons<span>.</span>
Kinetic Energy = 1/2×m×v²
50000 = 1/2×1000×v²
v = √100
v = 10 m/s