5cm 10cm
Q
Q2
Q3
Referring to the diagram above.
if Q1 = 9.0 uC, Q2 = -5.0 uC and Q3 = 32.0 uc,
what is the magnitude of the electric field at Q2, due to Q1 and Q3?

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

answer: its the third option.

step-by-step explanation:

i just took the test.

3+x

step-by-step explanation: