Answer:
False
Explanation:
Think of the electric potential in terms of potential energy. If you imagine a place with high elevation (A) and another one at sea level (B), a ball will roll from high potential to low potential (A-->B).
Everything in our universe wants to reach a lower state of energy if no external force is acted upon it. Every object tends to slow down (friction), a radioactive element dissipates energy (an unstable element releases energy to get to a stable state), water in the clouds comes down to the ground (rain experiencing difference in potential energy).
Electric potential is exactly the same, you just can't see it! It flows from higher voltage (which is a synonym for electric potential) to lower voltage.
By using an electric field, it is feasible to differentiate between these different forms of radiation.
<h3>What is a radioactive source?</h3>
A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.
The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.
Hence, by using an electric field, it is feasible to differentiate between these different forms of radiation.
To learn more about the radioactive source refer;
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The potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules
<h3>
What is the energy in a capacitor?</h3>
The energy stored in a capacitor is an electrostatic potential energy.
It is related to the charge(Q) and voltage (V) between the capacitor plates.
It is represented as 'U'.
<h3>
How to determine the potential difference</h3>
Formula:
Potential difference, V is the ratio of the charge to the capacitance of a capacitor.
It is calculated using:
V = Q ÷ C
Where Q = charge 5 × 10∧-5C and C = capacitance 10∧-9
Substitute the values into the equation
Potential difference, V = 5 × 10∧-5 ÷ 10∧-9 = 5 × 10∧4 volts
<h3>
How to determine the energy stored</h3>
Formula:
Energy, U = 1 ÷ 2 (QV)
Where Q= charge and V = potential difference across the capacitor
Energy, U = 1 ÷ 2 ( 5 × 10∧-5 × 5 × 10∧4)
= 0.5 × 25 × 10∧-1
= 0.5 × 2.5
= 1. 25 Joules
Therefore, the potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules
Learn more about capacitance here:
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