Answer:
0.235 nC
Step-by-step explanation:
Given:
= the magnitude of electric field =
![8.50\ kN/C =8.50* 10^(3)\ N/C](https://img.qammunity.org/2020/formulas/physics/college/md0xje1w6e2j7hh72biub5qy7nxvq4sujd.png)
= the magnitude of electric force on each antenna =
![2.00\ \mu N =2.00* 10^(-6)\ N](https://img.qammunity.org/2020/formulas/physics/college/5iq4jeqjm5nt7qmj0308ah852t7kjg2oyk.png)
= The magnitude of charge on each antenna
Since the electric field is the electric force applied on a charged body of unit charge.
![\therefore E = (F)/(q)\\\Rightarrow q =(F)/(E)\\\Rightarrow q =(2.00* 10^(-6)\ N)/(8.50* 10^(3)\ N/C)\\\Rightarrow q =0.235* 10^(-9)\ C\\\Rightarrow q =0.235\ nC](https://img.qammunity.org/2020/formulas/physics/college/jqlwat98md1oqzfr8cyyouo41cgbauwd2b.png)
Hence, the value of q is 0.235 nC.