Chapter 26 - The Electric Field - Conceptual Questions - Page 774: 2

See the figure below.

$$\color{blue}{\bf [a]}$$ Here the electric field will be zero at a point between the two point charges since the direction of the electric field is away from the positive charge. At the right from $+q$, the net electric field will always be toward the right. And at the left from $+4q$, the net electric field will always be toward the left. But between the two point charges, the electric field from $+4q$ is toward the right while the one from $+q$ is toward the left. $$E_{x,net}=E_{+4q}-E_{+q}=0$$ $$E_{x,net}=\dfrac{k(4q)}{ x ^2}-\dfrac{kq}{(3-x)^2}=0$$ Noting that the distance between the two point charges is 3 units. $$ \dfrac{4 }{ x ^2}-\dfrac{ 1}{(3-x)^2}=0$$ Solving for $x$ using any software calculator. So $x=2$ units or $x=6$ units which is dismissed since the distance between the two charges is only 3 units. Thus, at 2 units to the right from $+4q$, the net electric field is zero. See the first figure below. $$\color{blue}{\bf [b]}$$ Here we have two unlike charges. If we measured the direction of the electric field between these two charges where the negative charge $-q$ electric field is toward the right and the direction of the electric field of the positive charge $+4q$ is also toward the right as well. So the net electric field between the two point charges is toward the right and it can't be zero here. To the left from $+4q$, the direction of the electric filed of $-q$ is to the right while the direction of the electric field of $+4q$ is toward the left. But the electric field to the left from $+4q$ is weak for $-q$ for both reasons the direction and the small charge amount while the magnitude of the electric field of $+4q$ is always greater than the electric field of $-q$. Thus, the net electric field to the left from $+4q$ is toward the left and it can't be zero here. To the right from $-q$, the electric field can be zero at some point. $$E_{x,net}=E_{+4q}-E_{+q}=0$$ $$E_{x,net}=\dfrac{k(4q)}{ x ^2}-\dfrac{kq}{(3-x)^2}=0$$ Noting that the distance between the two point charges is 3 units. $$ \dfrac{4 }{ (x+3) ^2}-\dfrac{ 1}{(x)^2}=0$$ Solving for $x$ using any software calculator. So $x=-1$ which means to the left from $-q$ which is between the two charges so it must be dismissed, or $x=3$ units. Thus, at 3 units to the right from $-q$, the net electric field is zero. See the second figure below.