(a) Find the electric field at x = 5.00 cm in Figure 18.52(a), given that q = 1.00X10^-6;C . (b) At what position between 3.00 and 8.00 cm is the total electric field the same as that for –2q alone? (c) Can the electric field be zero anywhere between 0.00 and 8.00 cm? (d) At very large positive or negative values of x, the electric field approaches zero in both (a) and (b). In which does it most rapidly approach zero and why? (e) At what position to the right of 11.0 cm is the total electric field zero, other than at infinity? (Hint: A graphing calculator can yield considerable insight in this problem.)
Picture: q's are the particle charges, and the numbers are cm on the line.
A
-0---(+q)--5---(-2q)--10-(+q)---
B
-0-(-2q)----5---(+3q)--10----(-q)-
a.) I am having a problem with the signs, I know there should be 2 negative, and 1 positive. I also know that I should use Coulomb's law for each particle and then add them, but I'm not sure what to use for r.
b.) I'm confused about this question.
c.) I'm not sure how to approach this question, because there is no charge on 0.
d.) I understand :)
e.) Would I do the same tactic as C, but using different numbers? For example using (11+x)^2 for r in Coulomb's law.
I'm sorry for such a long question, but any help will be much appreciated! Thank you so much ahead of time!
You show a charge at x = 5 but you ask for the E field there
It is UNDEFINED. 1/r^2 as r--->0 is infinite
r is simply the distance between the charge and the location where you compute E
to the left of 5, the E field due to + q at 5 is left (repels)
however the E field due to the -2q at 10 is right (attract)
perhaps at some point they add to zero
say distance d left of 5
E = -k q/d^2 + 2 k q/ (d+5)^2
can that be zero?
q/d^2 = 2q/(d^2 + 10 d + 25)
2 d^2 = d^2 + 10 d + 25
d^2 - 10 d - 25 = 0
(d-5)(d-5) = 0
d = 5
so
five to the left of +5
or at zero the field is 0
d^2 -10 d - 25 = 0
d = [ 10 +/-sqrt (100+100) ]/2
= 5 +/- .5 sqrt (200)
= 5 +/- 5 sqrt 2
use +
so 5 sqrt 2 left of zero
= - 7.07
Ultimately, a dilemma that I'm passionate about. I have looked for data of this caliber for the previous various hours. Your internet site is greatly appreciated. bafbedbdabdd
It is rare for me to uncover something on the web that's as entertaining and intriguing as what you have got here. Your page is sweet, your graphics are great, and whats much more, you use source that are relevant to what youre saying. That you are undoubtedly one in a million, well done! ceefegcdkadd
Magnificent website. Lots of useful information here. Im sending it to some friends ans also sharing in delicious. And obviously, thanks for your sweat! eeggedfbkfkaadfd
a) To find the electric field at x = 5.00 cm, you need to calculate the electric field due to each charge and then add them up. The electric field due to a point charge q can be calculated using Coulomb's law:
Electric field (Ex) = k * q / r^2
Where:
- k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2)
- q is the charge of the particle
- r is the distance between the point where you want to find the electric field and the location of the charge
In this case, you have two charges, +q and -2q, located at 5.00 cm and 10.00 cm respectively. So, you need to calculate the electric field due to each charge separately and then add them up.
For the charge +q at x = 5.00 cm:
- q = 1.00 x 10^-6 C
- r = |x - 5.00 cm| = 0.00 cm (since the charge is located at x = 5.00 cm)
Hence, the electric field due to the +q charge is zero at x = 5.00 cm.
For the charge -2q at x = 10.00 cm:
- q = -2.00 x 1.00 x 10^-6 C
- r = |x - 10.00 cm| = 5.00 cm (since the charge is located at x = 10.00 cm and you want to find the electric field at x = 5.00 cm)
Using Coulomb's law, you can calculate the electric field due to -2q at 5.00 cm.
b) To find the position between 3.00 and 8.00 cm where the total electric field is the same as that for -2q alone, you need to identify the position where the electric field due to +q alone cancels out the electric field due to -q alone.
To do this, you would set up the equation:
Electric field due to +q = Electric field due to -q
You can calculate the individual electric fields due to +q and -q using Coulomb's law for the given positions. Then, set them equal to each other and solve for the position between 3.00 and 8.00 cm.
c) To determine if the electric field can be zero anywhere between 0.00 and 8.00 cm, you need to consider the positions and charges in the system.
In Figure 18.52(a), there are three charges: +q at x = 5.00 cm, -2q at x = 10.00 cm, and +q at x = 0.00 cm. If you calculate the electric field at different positions using Coulomb's law, you will see that the electric field is not zero for any position between 0.00 and 8.00 cm.
d) The electric field approaches zero at very large positive or negative values of x in both cases (a) and (b). This is because the electric field diminishes with distance, and as you move away from the charges, the distance becomes large and the electric field becomes weaker.
However, in terms of which case it most rapidly approaches zero, it depends on the charges and their positions. You would need to calculate the electric fields at very large positive or negative values of x for the given charges to compare and determine which one approaches zero more rapidly.
e) To find the position to the right of 11.0 cm where the total electric field is zero, you can use the same approach as in question (c), but with different numbers.
You would set up the equation:
Electric field due to +q1 + Electric field due to +q2 = Electric field due to -2q
Here, +q1 and +q2 represent the two charges located at x = 0.00 cm and x = 10.00 cm respectively. You can calculate the individual electric fields due to +q1 and +q2 at different positions and set them equal to the electric field due to -2q. Solve this equation to find the position to the right of 11.0 cm where the total electric field is zero.
Using a graphing calculator can provide additional insight and help visualize how the electric field varies with position.