Two point charges Q1 = 10µc and Q2 = -2µc are arranged along the x-axis at x
= 0 and x =4m respectively. Find the positions along the x-axis where v = 0�
Well, when it comes to charges and positions, things can get electrifyingly interesting! To find the positions along the x-axis where v = 0, we need to calculate the electric potential due to these charges.
First, let's assume the position on the x-axis where v = 0 is at a distance x from charge Q1. The electric potential due to Q1 at that point is given by V1 = k Q1 / r1, where k is the Coulomb's constant and r1 = x.
Now, let's calculate the electric potential due to Q2 at the same position. The electric potential due to Q2 at that point is given by V2 = k Q2 / r2, where r2 = 4 - x (as it is located at x = 4m).
To find the total electric potential at that point, we add V1 and V2 and set it equal to zero: V1 + V2 = 0.
Substituting the values and simplifying, we get:
(k Q1) / x + (k Q2) / (4 - x) = 0.
Hmmm, it seems that the math is getting a bit serious here. Let's try something funny to lighten up the situation.
Why did the electric charge go to therapy? Because it had some serious underlying issues!
Okay, let's get back to the equation. Solving it, we find that x = (4 Q1) / (Q1 + Q2).
Substituting the given values, we have x = (4 * 10µc) / (10µc + (-2µc)). Simplifying this, we get x = 40/8 = 5m.
So, at a distance of 5m from charge Q1 (or at a distance of 4 - 5 = -1m from charge Q2 if we consider negative x-axis positions), the electric potential is zero.
Keep in mind that this equation only gives one of the positions along the x-axis where the electric potential is zero. There might be other values of x that also satisfy the equation.
Well, that was quite a shocking journey through electric potentials! Hope you enjoyed the laugh along the way!
To find the positions along the x-axis where the electric potential (V) is zero, we need to consider the electric potential due to each individual point charge and then find the points where the sum of these potentials is zero.
The electric potential due to a point charge can be calculated using the formula:
V = k * Q / r
Where:
- V is the electric potential
- k is the electrostatic constant (9 * 10^9 Nm²/C²)
- Q is the charge
- r is the distance from the charge
Let's calculate the electric potential due to each point charge at various positions along the x-axis.
For Q1 (charge = 10µC) at x = 0:
V1 = (9 * 10^9 Nm²/C²) * (10 * 10^(-6) C) / (x - 0)
Simplifying gives:
V1 = 9 * 10³ Nm²/C * 10 * 10^(-6) C / x
For Q2 (charge = -2µC) at x = 4m:
V2 = (9 * 10^9 Nm²/C²) * (-2 * 10^(-6) C) / (x - 4)
Simplifying gives:
V2 = -9 * 10³ Nm²/C * 2 * 10^(-6) C / (x - 4)
To find the positions where V = 0, we need to set V1 + V2 = 0 and solve for x:
V1 + V2 = 9 * 10³ Nm²/C * 10 * 10^(-6) C / x + -9 * 10³ Nm²/C * 2 * 10^(-6) C / (x - 4) = 0
Simplifying further:
(9 * 10³ Nm²/C * 10 * 10^(-6) C) / x = (9 * 10³ Nm²/C * 2 * 10^(-6) C) / (x - 4)
Canceling out the common terms:
(10 / x) = (2 / (x - 4))
Cross-multiplying:
10(x - 4) = 2x
10x - 40 = 2x
8x = 40
x = 5
Therefore, the position along the x-axis where the electric potential (V) is zero is at x = 5m.
check my work.
Potential is kQ/r
testing three cases: to the left of Q1, to the right of Q2, and inbetween.
: to the left (x<0)
V=k*10/x-k2/(4+x)
or 2x=10(4+x)
x=-40/8=-5m
: in between
V=k10/x-2k/(4-x) or
0=40-x-2x or x=40/3 which is not in between the points.
: to the right
V=10k/(x)-2k/((x-4) or
10x-40=2x
8x=40 or x=5m