What is the magnitude of the field at point R?

[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.

 

[haruspex]

write expressions for the fields at points P and R in terms of q.

 

[paulimerci]

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?

 

[MatinSAR]

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]

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,

$$ 9\times 10^{-4} E_{P} = kq$$
$$ 9\times 10^{-4} E_{P} = E_R \times 100 \times 10^{-4}$$
$$ E_R = \frac {9 E_P} {100}$$

 

[MatinSAR]

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

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