please i need the answer in decimal not like this (1.02*10^3) i dont need it lik

Question

please i need the answer in decimal not like this (1.02*10^3) i dont need it like this i need it only numbers like 20.4 like this. thanks

Part (a) of the figure shows two charged particles fixed in place on an x axis with separation L. The ratio q1/q2 of their charge magnitudes is 4.00. Part (b) of the figure shows the x component E of their net electric field along the x axis just to the right of particle 2. The x axis scale is set by xs 30.9 cm. (a) At what value netX 19 of X 0 is E maximum? (b) If particle 2 has charge -4e, what is the value of that maximum? Assume e 1.602 x 10 net,X x (cm) (a) (a) Numbe Unit m (b) Number Unit N/C 2)

Explanation / Answer

The electric field surrounding a point charge q is given by
E(r) = kc·(q/r²) · e_r
kc = 1/(4··) = 8.988×109 Nm²/C² Coulomb's law
r distance from point charge to the point of evaluation
e_r unit vector point from the charge to the point off evaluation.

For the strength of the electrical field on the x-axis arising from a charge located at x_q on the the x-axis, this simplifies to:
E(x) = kc·q/(x-x_q)²
For our two charges:
E1(x) = kc·q1/(x-(-L))² = kc·q1/(x+L)²
E2(x) = kc·q2/(x-0)² = - kc·|q2|/x²
The total electrical field is the sum of the electrical field of the charges:
Enet(x) = E1(x) + E2(x)
= kc·q1/(x+L)² - kc·|q2|/x² = kc·|q2|·( (q1/|q2|)/(x+L)² - 1/x²)
= kc·|q2|·( 4/(x+L)² - 1/x²)

From the sketch (b) you find that the electrical field at x=20cm=0.2m is zero. With this information you can calculate L:
Enet(0.2cm) = 0
kc·|q2|·( 4/(0.2m+L)² - 1/x²) = 0
<=>
( 4/(0.2m+L)² - 1/x²) = 0
=>
L = 0.2m

(a)
At the maximum field strength the first derivative becomes zero:
dEnet/dx = 0
<=>
kc·|q2|·( -8/(x+L)³ + 2/x³) =0
<=>
(2/x² - 8/(x+L)³ + 2/x³) =0
<=>
xmax = L / (4^(1/3) - 1) = 0.2m / (4^(1/3) - 1)
= 0.34m = 34 cm

(b)
Enet,max = kc·|q2|·( 4/(xmax+L)² - 1/xmax²)
= 8.988×109 Nm²/C² · 3 · 1.602176487×10^-19C · ( 4/(0.34m + 0.2m)² - 1/(0.34m)²)
= 2.189×10^-8 V/m

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