The point along the line connecting the two charges is the total electric field due to the two charges equal to zero will be 0.3056.
The electric field strength is defined as the ratio of electric force and charge. The field's direction is determined by the direction of the force acting on the positive charge.
The megnitude of the electric field for charge 1 is;
[tex]\rm E_1 = (9 \times 10^9 )\times \frac{0.700 \times 10^{-9}}{x^2}[/tex]
The megnitude of the electric field for charge 1 is;
[tex]\rm E_2 = 9.0 \times 10^9 \times \frac{9.00 \times 10^{-9}}{(1.80-x)^2}[/tex]
The net electric field at point p is;
[tex]\rm \frac{0.700}{x^2} = \frac{9.00}{(1.40-x)^2} \\\\( \frac{1.40-x}{x} )^2 = \frac{9.00}{0.700} \\\\\ \frac{1.40-x}{x}=3.58 \\\\ \frac{140}{x} -1=3.58\\\\\ \frac{140}{x}=3.58+1\\\\ x=0.3056 \ m[/tex]
Hence, the point along the line connecting the two charges is the total electric field due to the two charges equal to zero will be 0.3056.
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