Where is the electric field is zero?

AI Thread Summary
The electric field between two charges of 1.5X10^-6 C and 3.0X10^-6 C, separated by 0.2 m, is zero at a point where the field strengths from both charges are equal. The equation derived from Coulomb's law shows that the electric field is neutral at a distance r from the smaller charge and (0.2 - r) from the larger charge. After algebraic manipulation, the solution indicates that r is approximately 0.08 m from the smaller charge. A mistake in calculations may lead to complex numbers, but following the correct algebraic steps and using the quadratic formula yields the expected result. The final answer confirms that the neutral point is indeed around 0.08 m from the smaller charge.
3.14159265358979
Messages
43
Reaction score
0
two charges of 1.5X10^-6 c and 3.0X10^-6 c are 0.2 m apart. Where is the electric field between them equal to zero?

heres what i got...

the electric field will be 0 when the field strength of the first charge minus the field strength of the second charge equals 0.

therefore,

q(1)-----x------P-------(0.2 - x)-------q(2)

where...
P is where the electric field equals 0
q(1) is the first charge
q(2) is the second charge
x is the distance (in metres) from the charge

[ kq(1) / (x)^2 ] - [ kq(2)/ ((0.2-x)^2) ] = 0

from here,
i cancel out the k's
find the common denominator and cancel it out once my numerator is expanded
try and use the quadratic equation to solve for x. however, when i try to solve for x i get a complex number...what am i doing wrong? the book says the answer is 0.08m (approx.) if you know a faster and much easier way, please do tell...thanks a bunch...
 
Physics news on Phys.org
Let the neutral point be a distance r from the smaller charge Q and a distance R-r from the bigger charge 2Q. Note that

kQ/r^2 - 2kQ/(R-r)^2 = 0 expresses the neutrality of the electric force along the line between the charges, by Coulomb's law.

After some algebraic manipulation you get

r^2 + 2Rr - R^2 = 0.

Applying the quadratic formula,

r = -R +/- sqrt(2)R. (I am too lazy to figure out how to stack the plus or minus symbol, so I wrote it as +/-.)

Discard the root that does not lie between the two charges. This leaves you with

r = [sqrt(2) - 1]R.

Here R=0.2 m, so you have

r = [1.414 - 1] 0.2

which works out to about 0.08 meters, that being the distance from the smaller charge.
 


Your approach is correct, but it seems like you may have made a mistake in your calculations when solving for x using the quadratic equation. Here is a step-by-step solution for finding the value of x:

1. Set up the equation:

[ kq(1) / (x)^2 ] - [ kq(2)/ ((0.2-x)^2) ] = 0

2. Cancel out the k's:

[ q(1) / (x)^2 ] - [ q(2)/ ((0.2-x)^2) ] = 0

3. Expand the denominators:

[ q(1) / x^2 ] - [ q(2) / (0.04 - 0.4x + x^2) ] = 0

4. Find the common denominator:

[ q(1)(0.04 - 0.4x + x^2) / x^2(0.04 - 0.4x + x^2) ] - [ q(2)(x^2) / x^2(0.04 - 0.4x + x^2) ] = 0

5. Simplify:

[ 0.04q(1) - 0.4q(1)x + q(1)x^2 - q(2)x^2 ] / [ x^2(0.04 - 0.4x + x^2) ] = 0

6. Combine like terms:

[ (q(1) - q(2))x^2 + 0.4q(1)x + 0.04q(1) ] / [ x^2(0.04 - 0.4x + x^2) ] = 0

7. Set the numerator equal to 0:

(q(1) - q(2))x^2 + 0.4q(1)x + 0.04q(1) = 0

8. Use the quadratic formula to solve for x:

x = [ -0.4q(1) ± √(0.4q(1))^2 - 4(q(1) - q(2))(0.04q(1)) ] / 2(q(1) - q(2))

9. Simplify:

x = [ -0.4q(1) ± √(
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top