Question

The electric potential (voltage) at a specific location is equal to the potential energy per unit charge a charged object would have if it were at that location. If the zero point of the voltage is at infinity, the numerical value of the voltage is equal to the numerical value of work done to bring in a unit charge from infinity to that location. Select Values and Grid in the menu, and drag one positive charge to the middle of the screen, right on top of two intersecting bold grid lines. Using the voltage meter, you should find that 1 m away from the charge, the voltage is 9 V. What is the voltage 2 m away from the charge?

Answer 1

The voltage 2 m away from the charged object is 36 V.

Step-by-step explanation:

The electric potential (voltage) at a distance of 1 m from the charged object is 9 V. To find the voltage at a distance of 2 m, we can use the inverse square law for electric potential:

V = kQ/r

Where V is the voltage, k is the electrostatic constant, Q is the charge, and r is the distance. Since the charge is unknown, we can use the ratio of the squares of the distances to find the voltage:

V₂/V₁ = (r₁/r₂)²

Substituting the given values, we have:

9/V₁ = (1/2)²

V₁ = 36 V

Therefore, the voltage 2 m away from the charge is 36 V.

Answer 2

4.5 V

Step-by-step explanation:

The electric potential produced around a single point charge is:

`V=k (q)/(r)`

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

We see that the electric potential is inversely proportional to the distance, r, so we can write:

`V_1 r_1 = V_2 r_2`

where

V1 = 9 V

r1 = 1 m

V2 = ?

r2 = 2 m

Solving the equation, we find the voltage 2 m away from the charge:

`V_2 = (V_1 r_1)/(r_2)=((9 V)(1 m))/(2 m)=4.5 V`