What is the magnitude of the field at point R?

• paulimerci
In summary, the conversation discusses solving a problem involving the electric field using the Pythagorean theorem and calculating the field at points R and P in terms of q. The correct ratio of ##\frac{E_R}{E_P}## is found to be ##\frac{9}{100}##.
paulimerci
Homework Statement
Problem attached below.
Relevant Equations
E = kq/r^2
I've no idea how to solve this problem. The sign of the charge is not mentioned, so I'm assuming the charge is "+". The charge exerts an outward electric field. Since two lengths of the right-angle triangle are given, I use the Pythagorean to find the hypotenuse, which is the distance between q and R, and it's found to be 10m.

$$E = \frac{kq}{r^2}$$
$$E = \frac{kq} {100}$$
I'm wondering why all the options have ##E_p## in the equation since it asks only for the magnitude of the field for point R. It would be great if anyone could explain how to solve this one.

Attachments

• Screenshot 2023-05-18 at 11.10.41 PM.png
16.4 KB · Views: 40
Last edited:
write expressions for the fields at points P and R in terms of q.

MatinSAR
haruspex said:
write expressions for the fields at points P and R in terms of q.
ok.
The electric field at point R is
$$E_{R} = \frac{kq}{100}$$
The electric field at point P is
$$E_{p} = \frac {kq}{9}$$
$$9\times E_{P} = kq$$
$$9\times E_{P} = E_R \times 100$$
$$E_R = \frac {9 E_P} {100}$$
Is it right?

paulimerci said:
The electric field at point R is
$$E_{R} = \frac{kq}{100}$$
The electric field at point P is
$$E_{p} = \frac {kq}{9}$$
These are wrong because you used centimeters in your calculations.
But final answer is ok since here we calculate the ratio.

paulimerci said:
Thank you for pointing out @MatinSAR. I edited it. Does it look ok now?
The electric field at point R is
$$E_{R} = \frac{kq}{100 \times 10^{-4}}$$
The electric field at point P is
$$E_{p} = \frac {kq}{9 \times 10^{-4}}$$
The ratio of ##\frac{E_P}{E_R}## gives,
paulimerci said:
$$9\times 10^{-4} E_{P} = kq$$
$$9\times 10^{-4} E_{P} = E_R \times 100 \times 10^{-4}$$
$$E_R = \frac {9 E_P} {100}$$

@paulimerci Yes, It's correct now.An easier way:

##\frac {E_R}{E_P}=\frac {\frac {kq}{r_R^2}}{\frac {kq}{r_P^2}}=(\frac {r_P}{r_R})^2=(\frac {3}{10})^2=\frac {9}{100}##

paulimerci

1. What is the definition of magnitude of a field?

The magnitude of a field refers to the strength or intensity of the field at a specific point. It is a measure of the force exerted by the field on a unit charge or mass at that point.

2. How is the magnitude of a field calculated?

The magnitude of a field is calculated using the formula: magnitude = force / charge or mass. This formula takes into account the strength of the field and the amount of charge or mass present at the point of interest.

3. What units are used to measure the magnitude of a field?

The units used to measure the magnitude of a field depend on the type of field being measured. For electric fields, the unit is newtons per coulomb (N/C). For magnetic fields, the unit is teslas (T). Other fields may have different units, such as gravitational fields which are measured in newtons per kilogram (N/kg).

4. How does the distance from the source affect the magnitude of a field?

The magnitude of a field decreases as the distance from the source increases. This is known as the inverse square law, which states that the strength of a field is inversely proportional to the square of the distance from the source. In other words, the farther you are from the source, the weaker the field will be.

5. Can the magnitude of a field be negative?

Yes, the magnitude of a field can be negative. This usually occurs when the field is a vector quantity, meaning it has both magnitude and direction. In this case, a negative magnitude indicates that the field is pointing in the opposite direction than a positive magnitude would.

• Introductory Physics Homework Help
Replies
5
Views
803
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
17
Views
982
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
890
• Introductory Physics Homework Help
Replies
32
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
178
• Introductory Physics Homework Help
Replies
11
Views
689
• Introductory Physics Homework Help
Replies
22
Views
1K