A positive charge magnitude q1=35.5 nc is located at origin. A negative charge q2= -4.5 nc is located on positive x axis at p= 16cm from origin.
Calculate the location on the x axis in cm where electric field is zero.
So far I know we set the eqns equal to each other
K*q1/(x^2)= K*q2/((16-x)^2)
To find the location on the x-axis where the electric field is zero, we can set the equations for the electric fields created by the two charges equal to each other. Let's start by setting up the equation:
$\frac{{K \cdot q_1}}{{x^2}} = \frac{{K \cdot q_2}}{{(16 - x)^2}}$
Here, K represents the electrostatic constant, $q_1$ and $q_2$ are the magnitudes of the charges, and x is the distance from the origin.
We can simplify this equation by canceling out the electrostatic constant, K:
$\frac{{q_1}}{{x^2}} = \frac{{q_2}}{{(16 - x)^2}}$
Given that $q_1 = 35.5 \, \text{nc}$ and $q_2 = -4.5 \, \text{nc}$, we can substitute these values into the equation:
$\frac{{35.5 \, \text{nc}}}{{x^2}} = \frac{{-4.5 \, \text{nc}}}{{(16 - x)^2}}$
Next, we can cross multiply to eliminate the fractions:
$35.5 \, \text{nc} \cdot (16 - x)^2 = -4.5 \, \text{nc} \cdot x^2$
Expanding and rearranging the equation gives us:
$35.5 \, \text{nc} \cdot (256 - 32x + x^2) = -4.5 \, \text{nc} \cdot x^2$
Simplifying further:
$9040 - 1132x + 35.5x^2 = -4.5x^2$
Combining like terms:
$40x^2 - 1132x + 9040 = 0$
To solve this quadratic equation, we can use the quadratic formula:
$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$
In our case, a = 40, b = -1132, and c = 9040. Plugging these values into the quadratic formula, we can solve for x.