Question

If the strength of the field is 3x10^6N/C and a proton is moved in the direction of electric field through a distance of 8mm.The change in potential of proton between the points A and B is
a) -1.5x10^-15V
b) 1.5x10^-15V
c) 2.4x10^4V
d) -2.4x10^4V​

Answer 1

Answer:

B

Explanation:

Hope it will helpful for you thanks..≥3≤

Answer 2

Answer: d)-2.4×10⁴V

Explanation:

Given that the electric field is uniform,

The potential between the two points will be equal to the product of the electric field and distance.

$\Delta V=-3\times 10^{6}N/C\times 8\times 10^{-3}m=-2.4 \times 10^{4}V$

For calculating the potential energy in this case:

$U=q\times V=1.6\times 10^{-19}C \times -2.4\times 10^{4}V\\U=-3.84\times 10^{-15}J$

Proton travels from high potential region to low potential region. So, Potential energy decreases.

Now we know that Potential difference can also be known if potential energy is known.

$V=U/q$