Two parallel plates of area 2.34-10* m?
have 7.07-107 C of charge placed on
them. A 6.62·105 C charge qi is placed
between the plates. What is the
magnitude of the electric force on qi?
Hint: How is force related to the field?
[?]N​

Two parallel plates of area 23410 mhave 707107 C of charge placed onthem A 662105 C charge qi is placedbetween the plates What is themagnitude of the electric f class=


Answer :

Sure! To find the value of $m$, we apply the angle bisector theorem, which states that the angle bisector divides the opposite side in proportion to the lengths of the other two sides.

Given the lines $y = x$ and $y = mx$, the slope of the line $y = x$ is 1, and the slope of the line $y = mx$ is $m$.

The angle bisector has a slope that is the geometric mean of the slopes of the lines that form the angle. The geometric mean of two numbers $a$ and $b$ is $\sqrt{ab}$.

So, the geometric mean of the slopes of the lines $y = x$ and $y = mx$ is $\sqrt{1 \cdot m} = \sqrt{m}$.

Now, we know that the line $y = \frac{1}{7}x$ bisects the angle between these two lines. Therefore, the slope of the line $y = \frac{1}{7}x$ is equal to the geometric mean of the slopes, which is $\sqrt{m}$.

Since the slope of the line $y = \frac{1}{7}x$ is $\frac{1}{7}$, we can solve for $m$ by setting $\sqrt{m}$ equal to $\frac{1}{7}$ and then squaring both sides to solve for $m$.

So, we have $\sqrt{m} = \frac{1}{7}$, and squaring both sides gives $m = \left(\frac{1}{7}\right)^2 = \frac{1}{49}$.

Therefore, the value of $m$ is $\frac{1}{49}$.

So, I made a mistake in my previous response, and I appreciate your patience. The correct value of $m$ is $\frac{1}{49}$, not $\frac{\sqrt{7}}{7}$.