The correct option is
D √2FThe force on test positive charge at the origin due to positive charge on the y-axis is acting downwards and due to the negative charge on the x-axis is acting towards the right as shown in the figure. Here the magnitude of both forces are equal that is equal to Fbecause F=kQqr and both case r and Qq are common. Thus, net force, R=√F2+F2=√2F
![532075_494740_ans_28bf4b4c9f584704b5a9550000fd4530.png](https://search-static.byjusweb.com/question-images/toppr_ext/questions/532075_494740_ans_28bf4b4c9f584704b5a9550000fd4530.png)