A uniform electric field of 10N/C exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 50cm.

To find the increase in electric potential when moving through a height of 50 cm in a uniform electric field of 10 N/C directed downward, we can use the relationship between electric potential difference (voltage) and electric field strength. The formula we’ll use is:

Understanding Electric Potential and Fields

The electric potential difference (ΔV) in a uniform electric field (E) is given by the equation:

ΔV = -E * d

Here, ΔV represents the change in electric potential, E is the magnitude of the electric field, and d is the displacement in the direction of the field. The negative sign indicates that when moving in the direction of the electric field, the potential decreases, while moving against it results in an increase in potential.

Identifying the Values

From the problem, we have:

• E = 10 N/C (the strength of the electric field)
• d = 0.5 m (the height through which we are moving, converted from 50 cm)

Calculating the Increase in Electric Potential

Since we are moving upwards, which is against the direction of the electric field, we should consider the negative sign in the formula. Plugging in the values, we get:

ΔV = -E * d

ΔV = - (10 N/C) * (0.5 m)

ΔV = -5 V

However, since we are interested in the increase in electric potential while moving upwards, we interpret the result as:

Increase in Electric Potential = 5 V

Visualizing the Concept

To help visualize this, think of the electric field as a slope going downward. If you were to walk uphill (upwards against the field), you would gain height and, correspondingly, your potential energy in the electric field would increase. In our case, moving up 50 cm in a 10 N/C electric field results in an increase of 5 volts in electric potential.

In summary, when you move against the direction of a uniform electric field, you gain electric potential, and for the given height of 50 cm in a 10 N/C field, the increase is 5 volts.