A and B are two points separated by a distance 5cm. Two charges $10\mu \mathrm{C}$ and $20\mu \mathrm{C}$ are placed at A and B. The resultant electric intensity at a point P outside the charges at a distance 5cm from $10\mu \mathrm{C}$ is

Given data: A and B separated by 5 cm.10μC and 20μC are at A and B.

Concept used: electric charges and fields 

Detailed solution:

Consider the scenario as it is presented to us. As indicated in the given figure, two charges are kept 5 cm apart , and we must determine the intensity of the electric field at a point p outside the system.

 E=14πε0Qr2 

 ⇒EAP=14πε010×10-65×10-22 

 ⇒EAP=9×109×10-525×10-4 

 ⇒EAP=925×108NC-1

We must determine the strength of the electric field at point P. Due to the charges at both sites A and B, we are certain that the point P will have an electric field. At point P, we may determine the specific electric field strengths resulting from each charge. At distance 5 am from p, the charge A of charge 10 C causes an electric field intensity at P that can be expressed as 

Similarly, the electric field at P due to charge B which is 10 cm away from P is given as 

 E=14πε0Qr2

 ⇒EBP=14πε020×10-610×10-22 

 ⇒EBP=9×109×2×10-5100×10-22

⇒EBP=9×109×2×10-5100×10-2

Now, we can find the total electric field at p as the sum of the fields due to A and B as

 EP=EAP+EBP 

 ⇒EP=925×108NC-1+18100×108NC-1

⇒EP=36+18100×108NC-1 

 ⇒EP=54×106NC-1