User:Thatguycalumb/sandbox

This is my sandbox.

$$v(t)=v_f+(v_i-v_f)e^{-t/RC}$$

$$e^{-t/RC}=\frac{v(t)-v_f}{v_i-v_f}$$

$$-t/RC=\ln\bigg(\frac{v(t)-v_f}{v_i-v_f}\bigg)$$

$$t=-RC\cdot\ln\bigg(\frac{v(t)-v_f}{v_i-v_f}\bigg)$$

$$t=RC\cdot\ln\Bigg(\frac{0-\frac{10}{3}}{0-\frac{5}{3}}\Bigg)$$

$$t=RC\cdot\ln\bigg(\frac{v_f-v_i}{v_f-v(t)}\bigg)$$

$$t=RC\cdot\ln\Bigg(\frac{5-\frac{5}{3}}{5-\frac{10}{3}}\Bigg)$$

$$t=RC\cdot\ln\Bigg(\frac{10/3}{5/3}\Bigg)$$

$$f=\frac{1}{(R+2R')(C)\cdot \ln 2}$$

$$f=\frac{1}{R'C\cdot\ln 2+(R+R')\cdot\ln 2}$$

$$\oint_s\mathbf{E} \cdot \mathrm{d} \mathbf{A} = \frac{Q_{enc}}{\varepsilon_0}$$

$$\binom{n}{k}=\frac{n!}{k!(n+k)!}$$

In electromagnetism, electric flux $$(\Phi_E)$$ is defined as the dot product of an electric field $$(\mathbf{E})$$ and the vector area $$(\mathbf{A})$$ of a surface it passes through.

$$\Phi_E = \mathbf{E} \cdot \mathbf{A}$$

In less formal terms, electric flux can be understood to represent the "amount of electric field" generated by a charge arrangement. If the electric field is not constant across a surface, the formula is better written as

$$\Phi_E = \iint_s \mathbf{E} \cdot \mathrm{d} \mathbf{A}$$

where $$s$$ represents a closed surface and $$\Phi_E $$ represents the net electric flux flowing through it. $$\Phi_E$$ is directly proportional to the net charge enclosed by the surface $$(Q_{enc})$$. This can be seen through Gauss' Law:

$$\oint_s \mathbf{E} \cdot \mathrm{d} \mathbf{A} = \frac{Q_{enc}}{\varepsilon_0}$$

where $$\varepsilon_0 $$ represents the electric constant $$(\sim 8.85 \times 10^{-12} \ \text{F}/\text{m})$$.

The SI unit for electric flux is the Volt metre (Vm).

$$\text{V} \cdot \text{m} = \frac{\text{N} \cdot \text{m}^2}{\text{C}} = \frac{\text{kg} \cdot \text{m}^3}{\text{s}^3 \cdot \text{A}}$$

$$- \pi \int \limits_4^8 \big(x-4\big)^2 \ \mathrm{d}x$$

$$-\pi \int \limits_4^8 \big(x^2-8x+16\big) \ \mathrm{d}x$$

$$-\pi \bigg[\frac{1}{3}x^3-4x^2+16x\bigg]_4^8$$

$$-\pi \bigg[\bigg(\frac{1}{3}(8)^3-4(8)^2+16(8)\bigg)-\bigg(\frac{1}{3}(4)^3-4(4)^2+16(4)\bigg)\bigg]$$

$$-\pi \bigg(\frac{128}{3}-\frac{64}{3}\bigg)$$

$$-\frac{64 \pi}{3}$$

$$- \pi \int \limits_4^8 \big(x-4\big)^2 \ \mathrm{d}x$$

$$- \pi \int \limits_{u(4)}^{u(8)} u^2 \ \mathrm{d}u$$

$$- \pi \bigg[\frac{1}{3}u^3\bigg]_0^4$$

$$- \frac{\pi}{3} \Big(4^3-0^3\Big)$$

$$-\frac{64 \pi}{3}$$

$$\frac{\sin x + x^2+\frac{5}{x}}{\sqrt{x+1} - \ln(x^2)} \cdot 2x^3$$

(sin(x) + x^2 + 5/x)/(sqrt(x+1)-ln(x^2))

$$((a) + (b) * (c)) / (d) *(e)$$

$$\frac{a + b \times c}{d} \times e$$

2 + (5(3 + a))