point a is located 0.25 m away from a charge of -2.1 x 10 -9 C. point b is located 0.50 m away from the charge. what is the electric potential difference Vb - Va between these two points?

The answer will be in Volts

k q[(1/X1) - (1/X2)], where

k = Coulomb's Law constant,
8.99 × 10^9 N*m^2/C^2
X1 = 0.50 m
X2 = 0.25 M
q = 2.1 x 10^-9 C

The answer will be in N*m or Joules

To find the electric potential difference (Vb - Va) between two points, we can use the equation:

Vb - Va = k * (Q / r1 - Q / r2)

Where:
- Vb - Va represents the electric potential difference between points b and a.
- k is Coulomb's constant, approximately equal to 9 x 10^9 N m²/C².
- Q is the charge of the particles.
- r1 and r2 are the distances from the charge to points a and b, respectively.

Given:
- Charge Q = -2.1 x 10^-9 C.
- Distance r1 = 0.25 m.
- Distance r2 = 0.50 m.

Plugging these values into the equation, we have:

Vb - Va = (9 x 10^9 N m²/C²) * (-2.1 x 10^-9 C / 0.25 m - (-2.1 x 10^-9 C / 0.50 m)

Simplifying the equation further:

Vb - Va = (9 x 10^9 N m²/C²) * (-2.1 x 10^-9 C * (1 / 0.25 m - 1 / 0.50 m))

Vb - Va = (9 x 10^9 N m²/C²) * (-2.1 x 10^-9 C * (4 - 2) / (0.50 m))

Vb - Va = (9 x 10^9 N m²/C²) * (-2.1 x 10^-9 C * 2 / (0.50 m))

Vb - Va = (9 x 10^9 N m²/C²) * (-4.2 x 10^-9 C / 0.50 m)

Now, calculating the numerical value:

Vb - Va = - (9 x 10^9 N m²/C²) * (4.2 x 10^-9 C / 0.50 m)

Vb - Va = - (9 x 10^9 N m²/C²) * (8.4 x 10^-9 C / m)

Vb - Va = -7.56 V

Therefore, the electric potential difference (Vb - Va) between points b and a is approximately -7.56 V.

To calculate the electric potential difference (Vb - Va) between two points (point b and point a), you can use the formula:

V = kQ/r

Where:
- V is the electric potential (in volts)
- k is the Coulomb's constant, approximately equal to 8.9875 x 10^9 N.m^2/C^2
- Q is the charge (in coulombs)
- r is the distance between the charge and the point (in meters)

Let's calculate the electric potential at point a (Va) and point b (Vb) separately, and then find the difference.

Given:
Charge (Q) = -2.1 x 10^-9 C
Distance from charge to point a (r_a) = 0.25 m
Distance from charge to point b (r_b) = 0.50 m

First, let's calculate the electric potential at point a (Va):
Va = (k * Q) / r_a

Substituting the given values:
Va = (8.9875 x 10^9 N.m^2/C^2 * -2.1 x 10^-9 C) / 0.25 m

Va ≈ -1.73 x 10^9 V

Similarly, let's calculate the electric potential at point b (Vb):
Vb = (k * Q) / r_b

Substituting the given values:
Vb = (8.9875 x 10^9 N.m^2/C^2 * -2.1 x 10^-9 C) / 0.50 m

Vb ≈ -8.68 x 10^8 V

Finally, to find the electric potential difference (Vb - Va):
Vb - Va ≈ -8.68 x 10^8 V - (-1.73 x 10^9 V)
Vb - Va ≈ -8.68 x 10^8 V + 1.73 x 10^9 V
Vb - Va ≈ 8.68 x 10^8 V

Therefore, the electric potential difference between point b and point a is approximately 8.68 x 10^8 volts.