Two point charges of 16 microcoulomb and - 9 microcoulomb are placed at8 cm distance. determine the position of the point at which resultant electric field is zero
It's wrong
Hello sir
When you took the square root u didint consider about the negative sign root 16 can be either +4/-4 right??
To determine the position at which the resultant electric field is zero between two point charges, we can use the principle of superposition. The superposition principle states that the net electric field at any point is the vector sum of the individual electric fields produced by each charge.
In this case, we have two point charges: +16 μC and -9 μC, separated by a distance of 8 cm. Let's denote the distance between the point of interest and the +16 μC charge as x.
To find the position at which the resultant electric field is zero, we need to find the distance x where the electric field produced by the +16 μC charge equals the electric field produced by the -9 μC charge, but with opposite direction.
The electric field produced by a point charge is given by the equation:
E = kq / r^2
Where:
E is the electric field,
k is the electrostatic constant (9 × 10^9 Nm²/C²),
q is the charge, and
r is the distance from the charge to the point of interest.
For the +16 μC charge:
E1 = (9 × 10^9 Nm²/C²) * (16 × 10^-6 C) / (x^2)
For the -9 μC charge:
E2 = (9 × 10^9 Nm²/C²) * (-9 × 10^-6 C) / ((8 - x)^2)
Since the resultant electric field is zero, E1 must be equal in magnitude and opposite in direction to E2. Thus, we can write the equation:
E1 = -E2
(9 × 10^9 Nm²/C²) * (16 × 10^-6 C) / (x^2) = - (9 × 10^9 Nm²/C²) * (-9 × 10^-6 C) / ((8 - x)^2)
Simplifying, we can cross-multiply to get:
(16 × 10^-6 C) / (x^2) = (-9 × 10^-6 C) / ((8 - x)^2)
Now, we can solve this equation for x.
Cross-multiplying again, we get:
(16 × 10^-6 C)*((8 - x)^2) = (-9 × 10^-6 C)*(x^2)
Expanding the squares, we have:
(16 × 10^-6 C)*(64 - 16x + x^2) = (-9 × 10^-6 C)*(x^2)
Multiplying both sides by 10^6 to remove the microcoulomb unit:
16*(64 - 16x + x^2) = -9*x^2
Simplifying further:
1024 - 256x + 16x^2 = -9x^2
Adding 9x^2 to both sides:
16x^2 - 256x + 9x^2 + 1024 = 0
25x^2 - 256x + 1024 = 0
Now, we can solve this quadratic equation to find the position of the point where the resultant electric field is zero. We can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 25, b = -256, and c = 1024.
Plugging in these values, we have:
x = (-(-256) ± √((-256)^2 - 4*25*1024)) / (2*25)
Simplifying:
x = (256 ± √(65536 - 12800)) / 50
x = (256 ± √(52736)) / 50
Calculating √(52736), we get:
x = (256 ± 229.6) / 50
Therefore, the two possible positions where the resultant electric field is zero are:
x1 = (256 + 229.6) / 50 = 9.72 cm (approximately)
x2 = (256 - 229.6) / 50 = 5.73 cm (approximately)
Hence, the position of the point at which the resultant electric field is zero between the two charges of 16 μC and -9 μC is approximately 9.72 cm and 5.73 cm from the +16 μC charge.
Let dist be x from charge -9micro coloumb
16|(8+×)^2=9|×^2
×|8+×=3|4
4×=3×+24
×=24
So dist of point where electric field is 0 is 24 cm from charge -9 micro coloumb